Mixing
Why do all elements commute in centre of an arbitrary finite group?
Clash Royale CLAN TAG#URR8PPP
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The section below is from wikipedia:
The center of G is always a subgroup of G. In particular:
1.
$Z(G)$ contains the identity element of $G$, $e$, because, for all $g$ in $G$, it commutes with $g$, $eg = g = ge$, by definition;
2.
If $x$ and $y$ are in $Z(G)$, then so is $xy$, by associativity: $(xy)g = x(yg) = x(gy) = (xg)y = (gx)y = g(xy)$ for each $g in G$; i.e., $Z(G)$ is closed;
3.
If $x$ is in $Z(G)$, then so is $x^−1$ as, for all $g$ in $G$, $x^−1$ commutes with $g$: $(gx = xg) implies (x^−1gxx^−1 = x^−1xgx^−1) implies (x^−1g = gx^−1)$.
Furthermore, the center of $G$ is always a normal subgroup of $G$, as it is closed under conjugation, as all elements commute.
I do not understand the last bit, why do all elements commute in $Z(G)$?
My attemt was to try to get $ab=ba$:
$a,b in Z(G) implies exists g,ga=ag, gb=bgimplies gagb=agbg$, but this doesnt't seem to lead anywhere useful.
abstract-algebra group-theory proof-writing finite-groups
asked 1 hour ago
zabop
1303
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zabop is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
2
down vote
favorite
The section below is from wikipedia:
The center of G is always a subgroup of G. In particular:
1.
$Z(G)$ contains the identity element of $G$, $e$, because, for all $g$ in $G$, it commutes with $g$, $eg = g = ge$, by definition;
2.
If $x$ and $y$ are in $Z(G)$, then so is $xy$, by associativity: $(xy)g = x(yg) = x(gy) = (xg)y = (gx)y = g(xy)$ for each $g in G$; i.e., $Z(G)$ is closed;
3.
If $x$ is in $Z(G)$, then so is $x^−1$ as, for all $g$ in $G$, $x^−1$ commutes with $g$: $(gx = xg) implies (x^−1gxx^−1 = x^−1xgx^−1) implies (x^−1g = gx^−1)$.
Furthermore, the center of $G$ is always a normal subgroup of $G$, as it is closed under conjugation, as all elements commute.
I do not understand the last bit, why do all elements commute in $Z(G)$?
My attemt was to try to get $ab=ba$:
$a,b in Z(G) implies exists g,ga=ag, gb=bgimplies gagb=agbg$, but this doesnt't seem to lead anywhere useful.
abstract-algebra group-theory proof-writing finite-groups
asked 1 hour ago
zabop
1303
New contributor
zabop is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Apply $g=b$.....
– Randall
1 hour ago
2
Your "$exists$" should be a "$forall$."
– Noah Schweber
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The section below is from wikipedia:
The center of G is always a subgroup of G. In particular:
1.
$Z(G)$ contains the identity element of $G$, $e$, because, for all $g$ in $G$, it commutes with $g$, $eg = g = ge$, by definition;
2.
If $x$ and $y$ are in $Z(G)$, then so is $xy$, by associativity: $(xy)g = x(yg) = x(gy) = (xg)y = (gx)y = g(xy)$ for each $g in G$; i.e., $Z(G)$ is closed;
3.
If $x$ is in $Z(G)$, then so is $x^−1$ as, for all $g$ in $G$, $x^−1$ commutes with $g$: $(gx = xg) implies (x^−1gxx^−1 = x^−1xgx^−1) implies (x^−1g = gx^−1)$.
Furthermore, the center of $G$ is always a normal subgroup of $G$, as it is closed under conjugation, as all elements commute.
I do not understand the last bit, why do all elements commute in $Z(G)$?
My attemt was to try to get $ab=ba$:
$a,b in Z(G) implies exists g,ga=ag, gb=bgimplies gagb=agbg$, but this doesnt't seem to lead anywhere useful.
abstract-algebra group-theory proof-writing finite-groups
asked 1 hour ago
zabop
1303
New contributor
zabop is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The section below is from wikipedia:
The center of G is always a subgroup of G. In particular:
1.
$Z(G)$ contains the identity element of $G$, $e$, because, for all $g$ in $G$, it commutes with $g$, $eg = g = ge$, by definition;
2.
If $x$ and $y$ are in $Z(G)$, then so is $xy$, by associativity: $(xy)g = x(yg) = x(gy) = (xg)y = (gx)y = g(xy)$ for each $g in G$; i.e., $Z(G)$ is closed;
3.
If $x$ is in $Z(G)$, then so is $x^−1$ as, for all $g$ in $G$, $x^−1$ commutes with $g$: $(gx = xg) implies (x^−1gxx^−1 = x^−1xgx^−1) implies (x^−1g = gx^−1)$.
Furthermore, the center of $G$ is always a normal subgroup of $G$, as it is closed under conjugation, as all elements commute.
I do not understand the last bit, why do all elements commute in $Z(G)$?
My attemt was to try to get $ab=ba$:
$a,b in Z(G) implies exists g,ga=ag, gb=bgimplies gagb=agbg$, but this doesnt't seem to lead anywhere useful.
abstract-algebra group-theory proof-writing finite-groups
abstract-algebra group-theory proof-writing finite-groups
asked 1 hour ago
zabop
1303
New contributor
zabop is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
zabop
1303
New contributor
zabop is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 1 hour ago
zabop
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asked 1 hour ago
zabop
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asked 1 hour ago
zabop
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1303
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zabop is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
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Apply $g=b$.....
– Randall
1 hour ago
2
Your "$exists$" should be a "$forall$."
– Noah Schweber
1 hour ago
add a comment |Â
Apply $g=b$.....
– Randall
1 hour ago
2
Your "$exists$" should be a "$forall$."
– Noah Schweber
1 hour ago
Apply $g=b$.....
– Randall
1 hour ago
Apply $g=b$.....
– Randall
1 hour ago
2
2
Your "$exists$" should be a "$forall$."
– Noah Schweber
1 hour ago
Your "$exists$" should be a "$forall$."
– Noah Schweber
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
6
down vote
accepted
The elements in the centre of a group, by definition, commute with all elements of the group $G$. Therefore they certainly commute with eachother.
Specifically, $Z(G) := zg = gz ; forall g in G$. If you have $a,b in Z(G)$, then putting $a$ and $b$ into the definition above gives you what you want, right?
answered 1 hour ago
Matt
2,391719
In my notes I have this: centre $Z$ is defined to be the set of elements which each form their own conjugacy class. From here, is it obvious that they commute with each other? I think what this means that there is at least one element in $G$ with which they commute. Why can we say that "they commute with all", not "at least one"?
– zabop
1 hour ago
Looking at your edit... I dont get why is it $Z(G) := zg = gz ; forall g in G$ and not $Z(G) := zg = gz ; exists g in G$
– zabop
1 hour ago
1
If an element $x in G$ forms its own conjugacy class, then you have that $ g in G = x$. In particular, $gxg^-1 = x ; forall g in G$. Thus, if $x in Z(G)$, $gx = xg$ for all $g in G$. My alternative definition then drops out, and things follow. Your assertion that "there is at least one element in $G$ with which they commute" is not quite correct, although it is understandable why you're confused. I think the definition which I give above of the centre of a group is a little easier for less experienced people to understand.
– Matt
1 hour ago
1
Additionally: It is worth you knowing that not only does your definition imply mine, but mine implies yours. They're equivalent. I've given you one implication, and you should see if you can get the reverse implication too.
– Matt
1 hour ago
add a comment |Â
up vote
0
down vote
Refer to @Matt's answer. The key step helping me to understand both the Q and that answer:
Two group elements $g_1, g_2 in G$ are conjugate to each other ($g_1 sim g_2$) if there exists any group element $g in G$ s.t.
$g_2 = gg_1g^−1$.
Now, if $g_1$ on its own conjugacy class $gg_1g^−1$ has always has to give $g_1$ for any $g$ because if it does not, then $g_1$ would be in the same conjugacy class as the result of $gg_1g^−1$, so for any $g$, we have $gg_1=g_1g$, if $g_1$ forms its own conjugacy class.
answered 52 mins ago
zabop
1303
New contributor
zabop is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
The elements in the centre of a group, by definition, commute with all elements of the group $G$. Therefore they certainly commute with eachother.
Specifically, $Z(G) := zg = gz ; forall g in G$. If you have $a,b in Z(G)$, then putting $a$ and $b$ into the definition above gives you what you want, right?
answered 1 hour ago
Matt
2,391719
In my notes I have this: centre $Z$ is defined to be the set of elements which each form their own conjugacy class. From here, is it obvious that they commute with each other? I think what this means that there is at least one element in $G$ with which they commute. Why can we say that "they commute with all", not "at least one"?
– zabop
1 hour ago
Looking at your edit... I dont get why is it $Z(G) := zg = gz ; forall g in G$ and not $Z(G) := zg = gz ; exists g in G$
– zabop
1 hour ago
1
If an element $x in G$ forms its own conjugacy class, then you have that $ g in G = x$. In particular, $gxg^-1 = x ; forall g in G$. Thus, if $x in Z(G)$, $gx = xg$ for all $g in G$. My alternative definition then drops out, and things follow. Your assertion that "there is at least one element in $G$ with which they commute" is not quite correct, although it is understandable why you're confused. I think the definition which I give above of the centre of a group is a little easier for less experienced people to understand.
– Matt
1 hour ago
1
Additionally: It is worth you knowing that not only does your definition imply mine, but mine implies yours. They're equivalent. I've given you one implication, and you should see if you can get the reverse implication too.
– Matt
1 hour ago
add a comment |Â
up vote
6
down vote
accepted
The elements in the centre of a group, by definition, commute with all elements of the group $G$. Therefore they certainly commute with eachother.
Specifically, $Z(G) := zg = gz ; forall g in G$. If you have $a,b in Z(G)$, then putting $a$ and $b$ into the definition above gives you what you want, right?
answered 1 hour ago
Matt
2,391719
In my notes I have this: centre $Z$ is defined to be the set of elements which each form their own conjugacy class. From here, is it obvious that they commute with each other? I think what this means that there is at least one element in $G$ with which they commute. Why can we say that "they commute with all", not "at least one"?
– zabop
1 hour ago
Looking at your edit... I dont get why is it $Z(G) := zg = gz ; forall g in G$ and not $Z(G) := zg = gz ; exists g in G$
– zabop
1 hour ago
1
If an element $x in G$ forms its own conjugacy class, then you have that $ g in G = x$. In particular, $gxg^-1 = x ; forall g in G$. Thus, if $x in Z(G)$, $gx = xg$ for all $g in G$. My alternative definition then drops out, and things follow. Your assertion that "there is at least one element in $G$ with which they commute" is not quite correct, although it is understandable why you're confused. I think the definition which I give above of the centre of a group is a little easier for less experienced people to understand.
– Matt
1 hour ago
1
Additionally: It is worth you knowing that not only does your definition imply mine, but mine implies yours. They're equivalent. I've given you one implication, and you should see if you can get the reverse implication too.
– Matt
1 hour ago
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
The elements in the centre of a group, by definition, commute with all elements of the group $G$. Therefore they certainly commute with eachother.
Specifically, $Z(G) := zg = gz ; forall g in G$. If you have $a,b in Z(G)$, then putting $a$ and $b$ into the definition above gives you what you want, right?
answered 1 hour ago
Matt
2,391719
The elements in the centre of a group, by definition, commute with all elements of the group $G$. Therefore they certainly commute with eachother.
Specifically, $Z(G) := zg = gz ; forall g in G$. If you have $a,b in Z(G)$, then putting $a$ and $b$ into the definition above gives you what you want, right?
answered 1 hour ago
Matt
2,391719
answered 1 hour ago
Matt
2,391719
answered 1 hour ago
Matt
2,391719
answered 1 hour ago
Matt
2,391719
2,391719
In my notes I have this: centre $Z$ is defined to be the set of elements which each form their own conjugacy class. From here, is it obvious that they commute with each other? I think what this means that there is at least one element in $G$ with which they commute. Why can we say that "they commute with all", not "at least one"?
– zabop
1 hour ago
Looking at your edit... I dont get why is it $Z(G) := zg = gz ; forall g in G$ and not $Z(G) := zg = gz ; exists g in G$
– zabop
1 hour ago
1
If an element $x in G$ forms its own conjugacy class, then you have that $ g in G = x$. In particular, $gxg^-1 = x ; forall g in G$. Thus, if $x in Z(G)$, $gx = xg$ for all $g in G$. My alternative definition then drops out, and things follow. Your assertion that "there is at least one element in $G$ with which they commute" is not quite correct, although it is understandable why you're confused. I think the definition which I give above of the centre of a group is a little easier for less experienced people to understand.
– Matt
1 hour ago
1
Additionally: It is worth you knowing that not only does your definition imply mine, but mine implies yours. They're equivalent. I've given you one implication, and you should see if you can get the reverse implication too.
– Matt
1 hour ago
add a comment |Â
In my notes I have this: centre $Z$ is defined to be the set of elements which each form their own conjugacy class. From here, is it obvious that they commute with each other? I think what this means that there is at least one element in $G$ with which they commute. Why can we say that "they commute with all", not "at least one"?
– zabop
1 hour ago
Looking at your edit... I dont get why is it $Z(G) := zg = gz ; forall g in G$ and not $Z(G) := zg = gz ; exists g in G$
– zabop
1 hour ago
1
If an element $x in G$ forms its own conjugacy class, then you have that $ g in G = x$. In particular, $gxg^-1 = x ; forall g in G$. Thus, if $x in Z(G)$, $gx = xg$ for all $g in G$. My alternative definition then drops out, and things follow. Your assertion that "there is at least one element in $G$ with which they commute" is not quite correct, although it is understandable why you're confused. I think the definition which I give above of the centre of a group is a little easier for less experienced people to understand.
– Matt
1 hour ago
1
Additionally: It is worth you knowing that not only does your definition imply mine, but mine implies yours. They're equivalent. I've given you one implication, and you should see if you can get the reverse implication too.
– Matt
1 hour ago
In my notes I have this: centre $Z$ is defined to be the set of elements which each form their own conjugacy class. From here, is it obvious that they commute with each other? I think what this means that there is at least one element in $G$ with which they commute. Why can we say that "they commute with all", not "at least one"?
– zabop
1 hour ago
In my notes I have this: centre $Z$ is defined to be the set of elements which each form their own conjugacy class. From here, is it obvious that they commute with each other? I think what this means that there is at least one element in $G$ with which they commute. Why can we say that "they commute with all", not "at least one"?
– zabop
1 hour ago
Looking at your edit... I dont get why is it $Z(G) := zg = gz ; forall g in G$ and not $Z(G) := zg = gz ; exists g in G$
– zabop
1 hour ago
Looking at your edit... I dont get why is it $Z(G) := zg = gz ; forall g in G$ and not $Z(G) := zg = gz ; exists g in G$
– zabop
1 hour ago
1
1
If an element $x in G$ forms its own conjugacy class, then you have that $ g in G = x$. In particular, $gxg^-1 = x ; forall g in G$. Thus, if $x in Z(G)$, $gx = xg$ for all $g in G$. My alternative definition then drops out, and things follow. Your assertion that "there is at least one element in $G$ with which they commute" is not quite correct, although it is understandable why you're confused. I think the definition which I give above of the centre of a group is a little easier for less experienced people to understand.
– Matt
1 hour ago
If an element $x in G$ forms its own conjugacy class, then you have that $ g in G = x$. In particular, $gxg^-1 = x ; forall g in G$. Thus, if $x in Z(G)$, $gx = xg$ for all $g in G$. My alternative definition then drops out, and things follow. Your assertion that "there is at least one element in $G$ with which they commute" is not quite correct, although it is understandable why you're confused. I think the definition which I give above of the centre of a group is a little easier for less experienced people to understand.
– Matt
1 hour ago
1
1
Additionally: It is worth you knowing that not only does your definition imply mine, but mine implies yours. They're equivalent. I've given you one implication, and you should see if you can get the reverse implication too.
– Matt
1 hour ago
Additionally: It is worth you knowing that not only does your definition imply mine, but mine implies yours. They're equivalent. I've given you one implication, and you should see if you can get the reverse implication too.
– Matt
1 hour ago
add a comment |Â
up vote
0
down vote
Refer to @Matt's answer. The key step helping me to understand both the Q and that answer:
Two group elements $g_1, g_2 in G$ are conjugate to each other ($g_1 sim g_2$) if there exists any group element $g in G$ s.t.
$g_2 = gg_1g^−1$.
Now, if $g_1$ on its own conjugacy class $gg_1g^−1$ has always has to give $g_1$ for any $g$ because if it does not, then $g_1$ would be in the same conjugacy class as the result of $gg_1g^−1$, so for any $g$, we have $gg_1=g_1g$, if $g_1$ forms its own conjugacy class.
answered 52 mins ago
zabop
1303
New contributor
zabop is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
Refer to @Matt's answer. The key step helping me to understand both the Q and that answer:
Two group elements $g_1, g_2 in G$ are conjugate to each other ($g_1 sim g_2$) if there exists any group element $g in G$ s.t.
$g_2 = gg_1g^−1$.
Now, if $g_1$ on its own conjugacy class $gg_1g^−1$ has always has to give $g_1$ for any $g$ because if it does not, then $g_1$ would be in the same conjugacy class as the result of $gg_1g^−1$, so for any $g$, we have $gg_1=g_1g$, if $g_1$ forms its own conjugacy class.
answered 52 mins ago
zabop
1303
New contributor
zabop is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Refer to @Matt's answer. The key step helping me to understand both the Q and that answer:
Two group elements $g_1, g_2 in G$ are conjugate to each other ($g_1 sim g_2$) if there exists any group element $g in G$ s.t.
$g_2 = gg_1g^−1$.
Now, if $g_1$ on its own conjugacy class $gg_1g^−1$ has always has to give $g_1$ for any $g$ because if it does not, then $g_1$ would be in the same conjugacy class as the result of $gg_1g^−1$, so for any $g$, we have $gg_1=g_1g$, if $g_1$ forms its own conjugacy class.
answered 52 mins ago
zabop
1303
New contributor
zabop is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Refer to @Matt's answer. The key step helping me to understand both the Q and that answer:
Two group elements $g_1, g_2 in G$ are conjugate to each other ($g_1 sim g_2$) if there exists any group element $g in G$ s.t.
$g_2 = gg_1g^−1$.
Now, if $g_1$ on its own conjugacy class $gg_1g^−1$ has always has to give $g_1$ for any $g$ because if it does not, then $g_1$ would be in the same conjugacy class as the result of $gg_1g^−1$, so for any $g$, we have $gg_1=g_1g$, if $g_1$ forms its own conjugacy class.
answered 52 mins ago
zabop
1303
New contributor
zabop is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 52 mins ago
zabop
1303
New contributor
zabop is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 52 mins ago
zabop
1303
answered 52 mins ago
zabop
1303
1303
New contributor
zabop is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
zabop is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
zabop is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
zabop is a new contributor. Be nice, and check out our Code of Conduct.
zabop is a new contributor. Be nice, and check out our Code of Conduct.
zabop is a new contributor. Be nice, and check out our Code of Conduct.
zabop is a new contributor. Be nice, and check out our Code of Conduct.
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Apply $g=b$.....
– Randall
1 hour ago
2
Your "$exists$" should be a "$forall$."
– Noah Schweber
1 hour ago