Mixing
Do a subgroup and quotient determine the original group?
Clash Royale CLAN TAG#URR8PPP
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I am wondering whether there is a counterexample which shows that subgroups and quotients don't determine the group.
More precisely, suppose there are two groups $G_1, G_2$ such that all of their proper non-trivial normal subgroups are 1-1 corresponding and if $1<H_1 < G_1, 1<H_2 < G_2$ are those proper normal subgroups that they correspond, then $H_1 simeq H_2$, and $G_1 / H_1 simeq G_2/H_2$. (Here $simeq$ means isomorphic.)
Then $G_1 simeq G_2$?
I guess it might be not true in general, but I don't know any nontrivial counterexample except the pair $(mathbbZ_p, mathbbZ_q)$.
Any comments on this will be highly appreciated!
group-theory finite-groups
asked Aug 8 at 8:03
user29422
41827
add a comment |Â
up vote
4
down vote
favorite
I am wondering whether there is a counterexample which shows that subgroups and quotients don't determine the group.
More precisely, suppose there are two groups $G_1, G_2$ such that all of their proper non-trivial normal subgroups are 1-1 corresponding and if $1<H_1 < G_1, 1<H_2 < G_2$ are those proper normal subgroups that they correspond, then $H_1 simeq H_2$, and $G_1 / H_1 simeq G_2/H_2$. (Here $simeq$ means isomorphic.)
Then $G_1 simeq G_2$?
I guess it might be not true in general, but I don't know any nontrivial counterexample except the pair $(mathbbZ_p, mathbbZ_q)$.
Any comments on this will be highly appreciated!
group-theory finite-groups
asked Aug 8 at 8:03
user29422
41827
Not to be annoying, but if $G_1/H_1 simeq G_2/H_2$ for any proper $H_1, H_2$, then you can take $H_1 = H_2 = 0$ and conclude, so you should exclude trivial subgroups from the hypothesis.
– Guido A.
Aug 8 at 8:07
1
Since this seems to have been interpreted differently in the two answer, could you clarify if this is what you intended: Let $A$ and $B$ be the sets of proper non-trivial subgroups of the groups and assume we have a bijective map $f: Ato B$ such that we for all $Hin A$ have $Hcong f(H)$ and $G_1/Hcong G_2/f(H)$ whenever $H$ is normal?
– Tobias Kildetoft
Aug 8 at 8:54
Following Tobias, what if the subgroups are not normal? Otherwise, why not simply take two simple groups?
– verret
Aug 8 at 9:09
1
The question seems clear enough to me, but you need to specify that normal subgroups correspond to normal subgroup in the correspondence between subgroups, since otherwise you cannot form quotient groups.
– Derek Holt
Aug 8 at 9:32
Thanks all. I corrected the original question as you commented.
– user29422
Aug 8 at 14:04
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I am wondering whether there is a counterexample which shows that subgroups and quotients don't determine the group.
More precisely, suppose there are two groups $G_1, G_2$ such that all of their proper non-trivial normal subgroups are 1-1 corresponding and if $1<H_1 < G_1, 1<H_2 < G_2$ are those proper normal subgroups that they correspond, then $H_1 simeq H_2$, and $G_1 / H_1 simeq G_2/H_2$. (Here $simeq$ means isomorphic.)
Then $G_1 simeq G_2$?
I guess it might be not true in general, but I don't know any nontrivial counterexample except the pair $(mathbbZ_p, mathbbZ_q)$.
Any comments on this will be highly appreciated!
group-theory finite-groups
asked Aug 8 at 8:03
user29422
41827
I am wondering whether there is a counterexample which shows that subgroups and quotients don't determine the group.
More precisely, suppose there are two groups $G_1, G_2$ such that all of their proper non-trivial normal subgroups are 1-1 corresponding and if $1<H_1 < G_1, 1<H_2 < G_2$ are those proper normal subgroups that they correspond, then $H_1 simeq H_2$, and $G_1 / H_1 simeq G_2/H_2$. (Here $simeq$ means isomorphic.)
Then $G_1 simeq G_2$?
I guess it might be not true in general, but I don't know any nontrivial counterexample except the pair $(mathbbZ_p, mathbbZ_q)$.
Any comments on this will be highly appreciated!
group-theory finite-groups
asked Aug 8 at 8:03
user29422
41827
edited Aug 8 at 14:02
asked Aug 8 at 8:03
user29422
41827
asked Aug 8 at 8:03
user29422
41827
asked Aug 8 at 8:03
user29422
41827
41827
Not to be annoying, but if $G_1/H_1 simeq G_2/H_2$ for any proper $H_1, H_2$, then you can take $H_1 = H_2 = 0$ and conclude, so you should exclude trivial subgroups from the hypothesis.
– Guido A.
Aug 8 at 8:07
1
Since this seems to have been interpreted differently in the two answer, could you clarify if this is what you intended: Let $A$ and $B$ be the sets of proper non-trivial subgroups of the groups and assume we have a bijective map $f: Ato B$ such that we for all $Hin A$ have $Hcong f(H)$ and $G_1/Hcong G_2/f(H)$ whenever $H$ is normal?
– Tobias Kildetoft
Aug 8 at 8:54
Following Tobias, what if the subgroups are not normal? Otherwise, why not simply take two simple groups?
– verret
Aug 8 at 9:09
1
The question seems clear enough to me, but you need to specify that normal subgroups correspond to normal subgroup in the correspondence between subgroups, since otherwise you cannot form quotient groups.
– Derek Holt
Aug 8 at 9:32
Thanks all. I corrected the original question as you commented.
– user29422
Aug 8 at 14:04
add a comment |Â
Not to be annoying, but if $G_1/H_1 simeq G_2/H_2$ for any proper $H_1, H_2$, then you can take $H_1 = H_2 = 0$ and conclude, so you should exclude trivial subgroups from the hypothesis.
– Guido A.
Aug 8 at 8:07
1
Since this seems to have been interpreted differently in the two answer, could you clarify if this is what you intended: Let $A$ and $B$ be the sets of proper non-trivial subgroups of the groups and assume we have a bijective map $f: Ato B$ such that we for all $Hin A$ have $Hcong f(H)$ and $G_1/Hcong G_2/f(H)$ whenever $H$ is normal?
– Tobias Kildetoft
Aug 8 at 8:54
Following Tobias, what if the subgroups are not normal? Otherwise, why not simply take two simple groups?
– verret
Aug 8 at 9:09
1
The question seems clear enough to me, but you need to specify that normal subgroups correspond to normal subgroup in the correspondence between subgroups, since otherwise you cannot form quotient groups.
– Derek Holt
Aug 8 at 9:32
Thanks all. I corrected the original question as you commented.
– user29422
Aug 8 at 14:04
Not to be annoying, but if $G_1/H_1 simeq G_2/H_2$ for any proper $H_1, H_2$, then you can take $H_1 = H_2 = 0$ and conclude, so you should exclude trivial subgroups from the hypothesis.
– Guido A.
Aug 8 at 8:07
Not to be annoying, but if $G_1/H_1 simeq G_2/H_2$ for any proper $H_1, H_2$, then you can take $H_1 = H_2 = 0$ and conclude, so you should exclude trivial subgroups from the hypothesis.
– Guido A.
Aug 8 at 8:07
1
1
Since this seems to have been interpreted differently in the two answer, could you clarify if this is what you intended: Let $A$ and $B$ be the sets of proper non-trivial subgroups of the groups and assume we have a bijective map $f: Ato B$ such that we for all $Hin A$ have $Hcong f(H)$ and $G_1/Hcong G_2/f(H)$ whenever $H$ is normal?
– Tobias Kildetoft
Aug 8 at 8:54
Since this seems to have been interpreted differently in the two answer, could you clarify if this is what you intended: Let $A$ and $B$ be the sets of proper non-trivial subgroups of the groups and assume we have a bijective map $f: Ato B$ such that we for all $Hin A$ have $Hcong f(H)$ and $G_1/Hcong G_2/f(H)$ whenever $H$ is normal?
– Tobias Kildetoft
Aug 8 at 8:54
Following Tobias, what if the subgroups are not normal? Otherwise, why not simply take two simple groups?
– verret
Aug 8 at 9:09
Following Tobias, what if the subgroups are not normal? Otherwise, why not simply take two simple groups?
– verret
Aug 8 at 9:09
1
1
The question seems clear enough to me, but you need to specify that normal subgroups correspond to normal subgroup in the correspondence between subgroups, since otherwise you cannot form quotient groups.
– Derek Holt
Aug 8 at 9:32
The question seems clear enough to me, but you need to specify that normal subgroups correspond to normal subgroup in the correspondence between subgroups, since otherwise you cannot form quotient groups.
– Derek Holt
Aug 8 at 9:32
Thanks all. I corrected the original question as you commented.
– user29422
Aug 8 at 14:04
Thanks all. I corrected the original question as you commented.
– user29422
Aug 8 at 14:04
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
9
down vote
accepted
There is a counterexample of order $605 = 11^2 times 5$ with the structure $11^2:5$. Let
$$G = langle x,y,z mid x^11=y^11=z^5=1, xy=yx, x^z=x^4, y^z=y^5 rangle,$$
$$H = langle x,y,z mid x^11=y^11=z^5=1, xy=yx, x^z=x^4, y^z=y^3 rangle.$$
These are $mathttSmallGroup(605,5)$ and $mathttSmallGroup(606,6)$ in the small groups database.
They both have $121$ conjugate subgroups of order $5$, two normal subgroups of order $11$ with nonabelian quotients of order $55$, $10$ non-normal subgroups of order $11$, $22$ non-normal subgroups of order $55$, and one normal subgroup of order $121$.
answered Aug 8 at 9:30
Derek Holt
50.1k53366
Aha Derek! That is the answer by the real master here. Excellent. +1 from me.
– Nicky Hekster
Aug 8 at 10:09
Wow! This is the answer what I wanted. Thank you very much!:)
– user29422
Aug 8 at 14:14
add a comment |Â
up vote
1
down vote
Smallest counterexample: $G_1=C_4$ generated by $x$ and $G_2=C_2 times C_2$. Take $H_1=langle x^2 rangle$ and $H_2=C_2 times 1$.
answered Aug 8 at 8:11
Nicky Hekster
27.1k53152
1
The proper subgroups of these do not correspond one-to-one
– Tobias Kildetoft
Aug 8 at 8:16
What does $C_n$ mean ??
– Anik Bhowmick
Aug 8 at 8:29
The cyclic group of order $n$ is denoted by $C_n$ (multiplicative notation) and this is also $mathbbZ/nmathbbZ$ (additive notation). But wait, Tobias triggered me: @user29422 what exaclty do you mean in your post by "such that all of their proper subgroups are 1-1 corresponding"?
– Nicky Hekster
Aug 8 at 8:49
1
I added a comment on the question asking for a clarification from the OP.
– Tobias Kildetoft
Aug 8 at 8:58
add a comment |Â
up vote
0
down vote
There are two (1) groups of order $4$. Each has a normal subgroup of order $2$ so the quotient groups are also order $2$. There is only one (1) group of order $2$ so the subgroups are isomorphic and the quotient groups are isomorphic.
(1) Up to isomorphism.
See The extension problem (Wikipedia)
answered Aug 8 at 8:11
badjohn
3,4551618
The OP asked for the proper subgroups to correspond one-to-one
– Tobias Kildetoft
Aug 8 at 8:16
1
@TobiasKildetoft Is there a technical meaning of "correspond" here that I am missing?
– badjohn
Aug 8 at 8:21
Not sure what that would be. The two groups of order $4$ have different numbers of subgroups, so they don't correspond.
– Tobias Kildetoft
Aug 8 at 8:22
@TobiasKildetoft I see what you mean now. So, it is a more interesting question than it first seemed.
– badjohn
Aug 8 at 8:26
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
There is a counterexample of order $605 = 11^2 times 5$ with the structure $11^2:5$. Let
$$G = langle x,y,z mid x^11=y^11=z^5=1, xy=yx, x^z=x^4, y^z=y^5 rangle,$$
$$H = langle x,y,z mid x^11=y^11=z^5=1, xy=yx, x^z=x^4, y^z=y^3 rangle.$$
These are $mathttSmallGroup(605,5)$ and $mathttSmallGroup(606,6)$ in the small groups database.
They both have $121$ conjugate subgroups of order $5$, two normal subgroups of order $11$ with nonabelian quotients of order $55$, $10$ non-normal subgroups of order $11$, $22$ non-normal subgroups of order $55$, and one normal subgroup of order $121$.
answered Aug 8 at 9:30
Derek Holt
50.1k53366
Aha Derek! That is the answer by the real master here. Excellent. +1 from me.
– Nicky Hekster
Aug 8 at 10:09
Wow! This is the answer what I wanted. Thank you very much!:)
– user29422
Aug 8 at 14:14
add a comment |Â
up vote
9
down vote
accepted
There is a counterexample of order $605 = 11^2 times 5$ with the structure $11^2:5$. Let
$$G = langle x,y,z mid x^11=y^11=z^5=1, xy=yx, x^z=x^4, y^z=y^5 rangle,$$
$$H = langle x,y,z mid x^11=y^11=z^5=1, xy=yx, x^z=x^4, y^z=y^3 rangle.$$
These are $mathttSmallGroup(605,5)$ and $mathttSmallGroup(606,6)$ in the small groups database.
They both have $121$ conjugate subgroups of order $5$, two normal subgroups of order $11$ with nonabelian quotients of order $55$, $10$ non-normal subgroups of order $11$, $22$ non-normal subgroups of order $55$, and one normal subgroup of order $121$.
answered Aug 8 at 9:30
Derek Holt
50.1k53366
Aha Derek! That is the answer by the real master here. Excellent. +1 from me.
– Nicky Hekster
Aug 8 at 10:09
Wow! This is the answer what I wanted. Thank you very much!:)
– user29422
Aug 8 at 14:14
add a comment |Â
up vote
9
down vote
accepted
up vote
9
down vote
accepted
There is a counterexample of order $605 = 11^2 times 5$ with the structure $11^2:5$. Let
$$G = langle x,y,z mid x^11=y^11=z^5=1, xy=yx, x^z=x^4, y^z=y^5 rangle,$$
$$H = langle x,y,z mid x^11=y^11=z^5=1, xy=yx, x^z=x^4, y^z=y^3 rangle.$$
These are $mathttSmallGroup(605,5)$ and $mathttSmallGroup(606,6)$ in the small groups database.
They both have $121$ conjugate subgroups of order $5$, two normal subgroups of order $11$ with nonabelian quotients of order $55$, $10$ non-normal subgroups of order $11$, $22$ non-normal subgroups of order $55$, and one normal subgroup of order $121$.
answered Aug 8 at 9:30
Derek Holt
50.1k53366
There is a counterexample of order $605 = 11^2 times 5$ with the structure $11^2:5$. Let
$$G = langle x,y,z mid x^11=y^11=z^5=1, xy=yx, x^z=x^4, y^z=y^5 rangle,$$
$$H = langle x,y,z mid x^11=y^11=z^5=1, xy=yx, x^z=x^4, y^z=y^3 rangle.$$
These are $mathttSmallGroup(605,5)$ and $mathttSmallGroup(606,6)$ in the small groups database.
They both have $121$ conjugate subgroups of order $5$, two normal subgroups of order $11$ with nonabelian quotients of order $55$, $10$ non-normal subgroups of order $11$, $22$ non-normal subgroups of order $55$, and one normal subgroup of order $121$.
answered Aug 8 at 9:30
Derek Holt
50.1k53366
answered Aug 8 at 9:30
Derek Holt
50.1k53366
answered Aug 8 at 9:30
Derek Holt
50.1k53366
answered Aug 8 at 9:30
Derek Holt
50.1k53366
50.1k53366
Aha Derek! That is the answer by the real master here. Excellent. +1 from me.
– Nicky Hekster
Aug 8 at 10:09
Wow! This is the answer what I wanted. Thank you very much!:)
– user29422
Aug 8 at 14:14
add a comment |Â
Aha Derek! That is the answer by the real master here. Excellent. +1 from me.
– Nicky Hekster
Aug 8 at 10:09
Wow! This is the answer what I wanted. Thank you very much!:)
– user29422
Aug 8 at 14:14
Aha Derek! That is the answer by the real master here. Excellent. +1 from me.
– Nicky Hekster
Aug 8 at 10:09
Aha Derek! That is the answer by the real master here. Excellent. +1 from me.
– Nicky Hekster
Aug 8 at 10:09
Wow! This is the answer what I wanted. Thank you very much!:)
– user29422
Aug 8 at 14:14
Wow! This is the answer what I wanted. Thank you very much!:)
– user29422
Aug 8 at 14:14
add a comment |Â
up vote
1
down vote
Smallest counterexample: $G_1=C_4$ generated by $x$ and $G_2=C_2 times C_2$. Take $H_1=langle x^2 rangle$ and $H_2=C_2 times 1$.
answered Aug 8 at 8:11
Nicky Hekster
27.1k53152
1
The proper subgroups of these do not correspond one-to-one
– Tobias Kildetoft
Aug 8 at 8:16
What does $C_n$ mean ??
– Anik Bhowmick
Aug 8 at 8:29
The cyclic group of order $n$ is denoted by $C_n$ (multiplicative notation) and this is also $mathbbZ/nmathbbZ$ (additive notation). But wait, Tobias triggered me: @user29422 what exaclty do you mean in your post by "such that all of their proper subgroups are 1-1 corresponding"?
– Nicky Hekster
Aug 8 at 8:49
1
I added a comment on the question asking for a clarification from the OP.
– Tobias Kildetoft
Aug 8 at 8:58
add a comment |Â
up vote
1
down vote
Smallest counterexample: $G_1=C_4$ generated by $x$ and $G_2=C_2 times C_2$. Take $H_1=langle x^2 rangle$ and $H_2=C_2 times 1$.
answered Aug 8 at 8:11
Nicky Hekster
27.1k53152
1
The proper subgroups of these do not correspond one-to-one
– Tobias Kildetoft
Aug 8 at 8:16
What does $C_n$ mean ??
– Anik Bhowmick
Aug 8 at 8:29
The cyclic group of order $n$ is denoted by $C_n$ (multiplicative notation) and this is also $mathbbZ/nmathbbZ$ (additive notation). But wait, Tobias triggered me: @user29422 what exaclty do you mean in your post by "such that all of their proper subgroups are 1-1 corresponding"?
– Nicky Hekster
Aug 8 at 8:49
1
I added a comment on the question asking for a clarification from the OP.
– Tobias Kildetoft
Aug 8 at 8:58
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Smallest counterexample: $G_1=C_4$ generated by $x$ and $G_2=C_2 times C_2$. Take $H_1=langle x^2 rangle$ and $H_2=C_2 times 1$.
answered Aug 8 at 8:11
Nicky Hekster
27.1k53152
Smallest counterexample: $G_1=C_4$ generated by $x$ and $G_2=C_2 times C_2$. Take $H_1=langle x^2 rangle$ and $H_2=C_2 times 1$.
answered Aug 8 at 8:11
Nicky Hekster
27.1k53152
answered Aug 8 at 8:11
Nicky Hekster
27.1k53152
answered Aug 8 at 8:11
Nicky Hekster
27.1k53152
answered Aug 8 at 8:11
Nicky Hekster
27.1k53152
27.1k53152
1
The proper subgroups of these do not correspond one-to-one
– Tobias Kildetoft
Aug 8 at 8:16
What does $C_n$ mean ??
– Anik Bhowmick
Aug 8 at 8:29
The cyclic group of order $n$ is denoted by $C_n$ (multiplicative notation) and this is also $mathbbZ/nmathbbZ$ (additive notation). But wait, Tobias triggered me: @user29422 what exaclty do you mean in your post by "such that all of their proper subgroups are 1-1 corresponding"?
– Nicky Hekster
Aug 8 at 8:49
1
I added a comment on the question asking for a clarification from the OP.
– Tobias Kildetoft
Aug 8 at 8:58
add a comment |Â
1
The proper subgroups of these do not correspond one-to-one
– Tobias Kildetoft
Aug 8 at 8:16
What does $C_n$ mean ??
– Anik Bhowmick
Aug 8 at 8:29
The cyclic group of order $n$ is denoted by $C_n$ (multiplicative notation) and this is also $mathbbZ/nmathbbZ$ (additive notation). But wait, Tobias triggered me: @user29422 what exaclty do you mean in your post by "such that all of their proper subgroups are 1-1 corresponding"?
– Nicky Hekster
Aug 8 at 8:49
1
I added a comment on the question asking for a clarification from the OP.
– Tobias Kildetoft
Aug 8 at 8:58
1
1
The proper subgroups of these do not correspond one-to-one
– Tobias Kildetoft
Aug 8 at 8:16
The proper subgroups of these do not correspond one-to-one
– Tobias Kildetoft
Aug 8 at 8:16
What does $C_n$ mean ??
– Anik Bhowmick
Aug 8 at 8:29
What does $C_n$ mean ??
– Anik Bhowmick
Aug 8 at 8:29
The cyclic group of order $n$ is denoted by $C_n$ (multiplicative notation) and this is also $mathbbZ/nmathbbZ$ (additive notation). But wait, Tobias triggered me: @user29422 what exaclty do you mean in your post by "such that all of their proper subgroups are 1-1 corresponding"?
– Nicky Hekster
Aug 8 at 8:49
The cyclic group of order $n$ is denoted by $C_n$ (multiplicative notation) and this is also $mathbbZ/nmathbbZ$ (additive notation). But wait, Tobias triggered me: @user29422 what exaclty do you mean in your post by "such that all of their proper subgroups are 1-1 corresponding"?
– Nicky Hekster
Aug 8 at 8:49
1
1
I added a comment on the question asking for a clarification from the OP.
– Tobias Kildetoft
Aug 8 at 8:58
I added a comment on the question asking for a clarification from the OP.
– Tobias Kildetoft
Aug 8 at 8:58
add a comment |Â
up vote
0
down vote
There are two (1) groups of order $4$. Each has a normal subgroup of order $2$ so the quotient groups are also order $2$. There is only one (1) group of order $2$ so the subgroups are isomorphic and the quotient groups are isomorphic.
(1) Up to isomorphism.
See The extension problem (Wikipedia)
answered Aug 8 at 8:11
badjohn
3,4551618
The OP asked for the proper subgroups to correspond one-to-one
– Tobias Kildetoft
Aug 8 at 8:16
1
@TobiasKildetoft Is there a technical meaning of "correspond" here that I am missing?
– badjohn
Aug 8 at 8:21
Not sure what that would be. The two groups of order $4$ have different numbers of subgroups, so they don't correspond.
– Tobias Kildetoft
Aug 8 at 8:22
@TobiasKildetoft I see what you mean now. So, it is a more interesting question than it first seemed.
– badjohn
Aug 8 at 8:26
add a comment |Â
up vote
0
down vote
There are two (1) groups of order $4$. Each has a normal subgroup of order $2$ so the quotient groups are also order $2$. There is only one (1) group of order $2$ so the subgroups are isomorphic and the quotient groups are isomorphic.
(1) Up to isomorphism.
See The extension problem (Wikipedia)
answered Aug 8 at 8:11
badjohn
3,4551618
The OP asked for the proper subgroups to correspond one-to-one
– Tobias Kildetoft
Aug 8 at 8:16
1
@TobiasKildetoft Is there a technical meaning of "correspond" here that I am missing?
– badjohn
Aug 8 at 8:21
Not sure what that would be. The two groups of order $4$ have different numbers of subgroups, so they don't correspond.
– Tobias Kildetoft
Aug 8 at 8:22
@TobiasKildetoft I see what you mean now. So, it is a more interesting question than it first seemed.
– badjohn
Aug 8 at 8:26
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There are two (1) groups of order $4$. Each has a normal subgroup of order $2$ so the quotient groups are also order $2$. There is only one (1) group of order $2$ so the subgroups are isomorphic and the quotient groups are isomorphic.
(1) Up to isomorphism.
See The extension problem (Wikipedia)
answered Aug 8 at 8:11
badjohn
3,4551618
There are two (1) groups of order $4$. Each has a normal subgroup of order $2$ so the quotient groups are also order $2$. There is only one (1) group of order $2$ so the subgroups are isomorphic and the quotient groups are isomorphic.
(1) Up to isomorphism.
See The extension problem (Wikipedia)
answered Aug 8 at 8:11
badjohn
3,4551618
answered Aug 8 at 8:11
badjohn
3,4551618
answered Aug 8 at 8:11
badjohn
3,4551618
answered Aug 8 at 8:11
badjohn
3,4551618
3,4551618
The OP asked for the proper subgroups to correspond one-to-one
– Tobias Kildetoft
Aug 8 at 8:16
1
@TobiasKildetoft Is there a technical meaning of "correspond" here that I am missing?
– badjohn
Aug 8 at 8:21
Not sure what that would be. The two groups of order $4$ have different numbers of subgroups, so they don't correspond.
– Tobias Kildetoft
Aug 8 at 8:22
@TobiasKildetoft I see what you mean now. So, it is a more interesting question than it first seemed.
– badjohn
Aug 8 at 8:26
add a comment |Â
The OP asked for the proper subgroups to correspond one-to-one
– Tobias Kildetoft
Aug 8 at 8:16
1
@TobiasKildetoft Is there a technical meaning of "correspond" here that I am missing?
– badjohn
Aug 8 at 8:21
Not sure what that would be. The two groups of order $4$ have different numbers of subgroups, so they don't correspond.
– Tobias Kildetoft
Aug 8 at 8:22
@TobiasKildetoft I see what you mean now. So, it is a more interesting question than it first seemed.
– badjohn
Aug 8 at 8:26
The OP asked for the proper subgroups to correspond one-to-one
– Tobias Kildetoft
Aug 8 at 8:16
The OP asked for the proper subgroups to correspond one-to-one
– Tobias Kildetoft
Aug 8 at 8:16
1
1
@TobiasKildetoft Is there a technical meaning of "correspond" here that I am missing?
– badjohn
Aug 8 at 8:21
@TobiasKildetoft Is there a technical meaning of "correspond" here that I am missing?
– badjohn
Aug 8 at 8:21
Not sure what that would be. The two groups of order $4$ have different numbers of subgroups, so they don't correspond.
– Tobias Kildetoft
Aug 8 at 8:22
Not sure what that would be. The two groups of order $4$ have different numbers of subgroups, so they don't correspond.
– Tobias Kildetoft
Aug 8 at 8:22
@TobiasKildetoft I see what you mean now. So, it is a more interesting question than it first seemed.
– badjohn
Aug 8 at 8:26
@TobiasKildetoft I see what you mean now. So, it is a more interesting question than it first seemed.
– badjohn
Aug 8 at 8:26
add a comment |Â
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Not to be annoying, but if $G_1/H_1 simeq G_2/H_2$ for any proper $H_1, H_2$, then you can take $H_1 = H_2 = 0$ and conclude, so you should exclude trivial subgroups from the hypothesis.
– Guido A.
Aug 8 at 8:07
1
Since this seems to have been interpreted differently in the two answer, could you clarify if this is what you intended: Let $A$ and $B$ be the sets of proper non-trivial subgroups of the groups and assume we have a bijective map $f: Ato B$ such that we for all $Hin A$ have $Hcong f(H)$ and $G_1/Hcong G_2/f(H)$ whenever $H$ is normal?
– Tobias Kildetoft
Aug 8 at 8:54
Following Tobias, what if the subgroups are not normal? Otherwise, why not simply take two simple groups?
– verret
Aug 8 at 9:09
1
The question seems clear enough to me, but you need to specify that normal subgroups correspond to normal subgroup in the correspondence between subgroups, since otherwise you cannot form quotient groups.
– Derek Holt
Aug 8 at 9:32
Thanks all. I corrected the original question as you commented.
– user29422
Aug 8 at 14:04