Prove that this is a surjection and find the kernel
Clash Royale CLAN TAG#URR8PPP
$begingroup$
We have a map $alpha: G rightarrow S(G)$, where $S(G)$ is the group of all bijections from $G$ to $G$. And $alpha(g) = f_g$, where $f_g(a) = gag^-1$. It's easy to prove that this is a homomorphism and its kernel is the set of $gin G$ such that $f_g = operatornameId_G$, i.e. $f_g(a) = gag^-1 = a$ $forall ain G$. It means that $ga = ag$ $forall ain G$ and it is the definition of $Z(G)$, the center of group $G$.
But what about proving that this is a surjection?
I find it obvious by definition (I mean that $alpha(f_g) = g$) But how to prove that this is a surjection strongly and what about my solution of kernel? Is it ok?
abstract-algebra group-theory
edited Mar 10 at 1:27
J. W. Tanner
4,5041320
asked Mar 9 at 12:56
ErlGreyErlGrey
537
$endgroup$
add a comment |
$begingroup$
We have a map $alpha: G rightarrow S(G)$, where $S(G)$ is the group of all bijections from $G$ to $G$. And $alpha(g) = f_g$, where $f_g(a) = gag^-1$. It's easy to prove that this is a homomorphism and its kernel is the set of $gin G$ such that $f_g = operatornameId_G$, i.e. $f_g(a) = gag^-1 = a$ $forall ain G$. It means that $ga = ag$ $forall ain G$ and it is the definition of $Z(G)$, the center of group $G$.
But what about proving that this is a surjection?
I find it obvious by definition (I mean that $alpha(f_g) = g$) But how to prove that this is a surjection strongly and what about my solution of kernel? Is it ok?
abstract-algebra group-theory
edited Mar 10 at 1:27
J. W. Tanner
4,5041320
asked Mar 9 at 12:56
ErlGreyErlGrey
537
$endgroup$
$begingroup$
It will not be a surjection in general. For a group of order $3$, $S (G) $ is a group of order $6$.
$endgroup$
– Thomas Shelby
Mar 9 at 13:02
add a comment |
$begingroup$
We have a map $alpha: G rightarrow S(G)$, where $S(G)$ is the group of all bijections from $G$ to $G$. And $alpha(g) = f_g$, where $f_g(a) = gag^-1$. It's easy to prove that this is a homomorphism and its kernel is the set of $gin G$ such that $f_g = operatornameId_G$, i.e. $f_g(a) = gag^-1 = a$ $forall ain G$. It means that $ga = ag$ $forall ain G$ and it is the definition of $Z(G)$, the center of group $G$.
But what about proving that this is a surjection?
I find it obvious by definition (I mean that $alpha(f_g) = g$) But how to prove that this is a surjection strongly and what about my solution of kernel? Is it ok?
abstract-algebra group-theory
edited Mar 10 at 1:27
J. W. Tanner
4,5041320
asked Mar 9 at 12:56
ErlGreyErlGrey
537
$endgroup$
We have a map $alpha: G rightarrow S(G)$, where $S(G)$ is the group of all bijections from $G$ to $G$. And $alpha(g) = f_g$, where $f_g(a) = gag^-1$. It's easy to prove that this is a homomorphism and its kernel is the set of $gin G$ such that $f_g = operatornameId_G$, i.e. $f_g(a) = gag^-1 = a$ $forall ain G$. It means that $ga = ag$ $forall ain G$ and it is the definition of $Z(G)$, the center of group $G$.
But what about proving that this is a surjection?
I find it obvious by definition (I mean that $alpha(f_g) = g$) But how to prove that this is a surjection strongly and what about my solution of kernel? Is it ok?
abstract-algebra group-theory
abstract-algebra group-theory
edited Mar 10 at 1:27
J. W. Tanner
4,5041320
asked Mar 9 at 12:56
ErlGreyErlGrey
537
edited Mar 10 at 1:27
J. W. Tanner
4,5041320
asked Mar 9 at 12:56
ErlGreyErlGrey
537
edited Mar 10 at 1:27
J. W. Tanner
4,5041320
edited Mar 10 at 1:27
J. W. Tanner
4,5041320
edited Mar 10 at 1:27
J. W. Tanner
4,5041320
4,5041320
asked Mar 9 at 12:56
ErlGreyErlGrey
537
asked Mar 9 at 12:56
ErlGreyErlGrey
537
asked Mar 9 at 12:56
ErlGreyErlGrey
537
537
$begingroup$
It will not be a surjection in general. For a group of order $3$, $S (G) $ is a group of order $6$.
$endgroup$
– Thomas Shelby
Mar 9 at 13:02
add a comment |
$begingroup$
It will not be a surjection in general. For a group of order $3$, $S (G) $ is a group of order $6$.
$endgroup$
– Thomas Shelby
Mar 9 at 13:02
$begingroup$
It will not be a surjection in general. For a group of order $3$, $S (G) $ is a group of order $6$.
$endgroup$
– Thomas Shelby
Mar 9 at 13:02
$begingroup$
It will not be a surjection in general. For a group of order $3$, $S (G) $ is a group of order $6$.
$endgroup$
– Thomas Shelby
Mar 9 at 13:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In general, $alpha$ is not surjective. This follows already from the fact that $|S(G)|=n!$ when $|G|=n$ (and $n!>n$ for $n>2$). Also, we have $alpha(g)(1)=1$ for all $gin G$, but (unless $n=1$) there exist bijections $in S(G)$ that map $1$ elsewhere.
In summary, $alpha$ is surjective iff $G$ is trivial.
answered Mar 9 at 13:02
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
$begingroup$
Thank you so much! I have not thought about this obvious observation!
$endgroup$
– ErlGrey
Mar 9 at 13:17
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In general, $alpha$ is not surjective. This follows already from the fact that $|S(G)|=n!$ when $|G|=n$ (and $n!>n$ for $n>2$). Also, we have $alpha(g)(1)=1$ for all $gin G$, but (unless $n=1$) there exist bijections $in S(G)$ that map $1$ elsewhere.
In summary, $alpha$ is surjective iff $G$ is trivial.
answered Mar 9 at 13:02
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
$begingroup$
Thank you so much! I have not thought about this obvious observation!
$endgroup$
– ErlGrey
Mar 9 at 13:17
add a comment |
$begingroup$
In general, $alpha$ is not surjective. This follows already from the fact that $|S(G)|=n!$ when $|G|=n$ (and $n!>n$ for $n>2$). Also, we have $alpha(g)(1)=1$ for all $gin G$, but (unless $n=1$) there exist bijections $in S(G)$ that map $1$ elsewhere.
In summary, $alpha$ is surjective iff $G$ is trivial.
answered Mar 9 at 13:02
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
$begingroup$
Thank you so much! I have not thought about this obvious observation!
$endgroup$
– ErlGrey
Mar 9 at 13:17
add a comment |
$begingroup$
In general, $alpha$ is not surjective. This follows already from the fact that $|S(G)|=n!$ when $|G|=n$ (and $n!>n$ for $n>2$). Also, we have $alpha(g)(1)=1$ for all $gin G$, but (unless $n=1$) there exist bijections $in S(G)$ that map $1$ elsewhere.
In summary, $alpha$ is surjective iff $G$ is trivial.
answered Mar 9 at 13:02
Hagen von EitzenHagen von Eitzen
283k23273508
$endgroup$
In general, $alpha$ is not surjective. This follows already from the fact that $|S(G)|=n!$ when $|G|=n$ (and $n!>n$ for $n>2$). Also, we have $alpha(g)(1)=1$ for all $gin G$, but (unless $n=1$) there exist bijections $in S(G)$ that map $1$ elsewhere.
In summary, $alpha$ is surjective iff $G$ is trivial.
answered Mar 9 at 13:02
Hagen von EitzenHagen von Eitzen
283k23273508
answered Mar 9 at 13:02
Hagen von EitzenHagen von Eitzen
283k23273508
answered Mar 9 at 13:02
Hagen von EitzenHagen von Eitzen
283k23273508
answered Mar 9 at 13:02
Hagen von EitzenHagen von Eitzen
283k23273508
283k23273508
$begingroup$
Thank you so much! I have not thought about this obvious observation!
$endgroup$
– ErlGrey
Mar 9 at 13:17
add a comment |
$begingroup$
Thank you so much! I have not thought about this obvious observation!
$endgroup$
– ErlGrey
Mar 9 at 13:17
$begingroup$
Thank you so much! I have not thought about this obvious observation!
$endgroup$
– ErlGrey
Mar 9 at 13:17
$begingroup$
Thank you so much! I have not thought about this obvious observation!
$endgroup$
– ErlGrey
Mar 9 at 13:17
add a comment |
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$begingroup$
It will not be a surjection in general. For a group of order $3$, $S (G) $ is a group of order $6$.
$endgroup$
– Thomas Shelby
Mar 9 at 13:02