Prove that this is a surjection and find the kernel
$begingroup$
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 isa homomorphism and its kernel is the set of $gin G$ such that $f_g = operatorname{Id}_G$, 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 43 mins ago
Bernard
122k741116
asked 53 mins ago
ErlGreyErlGrey
224
$endgroup$
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 isa homomorphism and its kernel is the set of $gin G$ such that $f_g = operatorname{Id}_G$, 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 43 mins ago
Bernard
122k741116
asked 53 mins ago
ErlGreyErlGrey
224
$endgroup$
$begingroup$
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
47 mins ago
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 isa homomorphism and its kernel is the set of $gin G$ such that $f_g = operatorname{Id}_G$, 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 43 mins ago
Bernard
122k741116
asked 53 mins ago
ErlGreyErlGrey
224
$endgroup$
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 isa homomorphism and its kernel is the set of $gin G$ such that $f_g = operatorname{Id}_G$, 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 43 mins ago
Bernard
122k741116
asked 53 mins ago
ErlGreyErlGrey
224
edited 43 mins ago
Bernard
122k741116
asked 53 mins ago
ErlGreyErlGrey
224
edited 43 mins ago
Bernard
122k741116
edited 43 mins ago
Bernard
122k741116
edited 43 mins ago
Bernard
122k741116
122k741116
asked 53 mins ago
ErlGreyErlGrey
224
asked 53 mins ago
ErlGreyErlGrey
224
asked 53 mins ago
ErlGreyErlGrey
224
224
$begingroup$
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
47 mins ago
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
47 mins ago
$begingroup$
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
47 mins ago
$begingroup$
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
47 mins ago
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 46 mins ago
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
$begingroup$
Thank you so much! I have not thought about this obvious observation!
$endgroup$
– ErlGrey
32 mins ago
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 46 mins ago
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
$begingroup$
Thank you so much! I have not thought about this obvious observation!
$endgroup$
– ErlGrey
32 mins ago
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 46 mins ago
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
$begingroup$
Thank you so much! I have not thought about this obvious observation!
$endgroup$
– ErlGrey
32 mins ago
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 46 mins ago
Hagen von EitzenHagen von Eitzen
282k23272507
$endgroup$
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 46 mins ago
Hagen von EitzenHagen von Eitzen
282k23272507
answered 46 mins ago
Hagen von EitzenHagen von Eitzen
282k23272507
answered 46 mins ago
Hagen von EitzenHagen von Eitzen
282k23272507
answered 46 mins ago
Hagen von EitzenHagen von Eitzen
282k23272507
282k23272507
$begingroup$
Thank you so much! I have not thought about this obvious observation!
$endgroup$
– ErlGrey
32 mins ago
add a comment |
$begingroup$
Thank you so much! I have not thought about this obvious observation!
$endgroup$
– ErlGrey
32 mins ago
$begingroup$
Thank you so much! I have not thought about this obvious observation!
$endgroup$
– ErlGrey
32 mins ago
$begingroup$
Thank you so much! I have not thought about this obvious observation!
$endgroup$
– ErlGrey
32 mins ago
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
47 mins ago