# group homomorphism

• Jan 15th 2010, 02:41 PM
Icarus0
group homomorphism
Let G be a group. Is it true that there is only one surjective group homomorphism from G to {-1,1}? If this is true how can I proof it?

I need this to proof that f:Sym(n) -> {-1,1} is sgn if f is multiplicative and surjective. Obviously f and sgn are surjective group homomorphisms...
• Jan 15th 2010, 09:27 PM
Amer
Quote:

Originally Posted by Icarus0
Let G be a group. Is it true that there is only one surjective group homomorphism from G to {-1,1}? If this is true how can I proof it?

I need this to proof that f:Sym(n) -> {-1,1} is sgn if f is multiplicative and surjective. Obviously f and sgn are surjective group homomorphisms...

what do you mean by Sym(n)
you want to prove that f is sgn what do you mean by sgn
write what you want without using shortcuts so all can understand your question
• Jan 15th 2010, 09:32 PM
NonCommAlg
Quote:

Originally Posted by Icarus0
Let G be a group. Is it true that there is only one surjective group homomorphism from G to {-1,1}? If this is true how can I proof it?

I need this to proof that f:Sym(n) -> {-1,1} is sgn if f is multiplicative and surjective. Obviously f and sgn are surjective group homomorphisms...

well, since $f$ is surjective, we have $[S_n: \ker f]= 2$ but we know (i hope you know!) that $A_n$ is the unique subgroup of $S_n$ with index 2. thus $\ker f = A_n$ and you're done.
• Jan 16th 2010, 12:18 AM
tonio
Quote:

Originally Posted by Icarus0
Let G be a group. Is it true that there is only one surjective group homomorphism from G to {-1,1}? If this is true how can I proof it?

No, it isn't. If there exists such an onto homomorphism then the group has to have even order...and not all (finite) groups have even order, of course.

Tonio

I need this to proof that f:Sym(n) -> {-1,1} is sgn if f is multiplicative and surjective. Obviously f and sgn are surjective group homomorphisms...

.
• Jan 16th 2010, 04:47 AM
Icarus0
Quote:

Originally Posted by Amer
what do you mean by Sym(n)
you want to prove that f is sgn what do you mean by sgn
write what you want without using shortcuts so all can understand your question

Sym(n) is the group that consists of all permutations of the first n natural numbers. sgn is the sign of the permutation.

I don't understand anything... S_n, A_n???

Quote:

No, it isn't. If there exists such an onto homomorphism then the group has to have even order...and not all (finite) groups have even order, of course.

Tonio
Has sym(n) even order?
• Jan 16th 2010, 08:40 AM
tonio
Quote:

Originally Posted by Icarus0
Sym(n) is the group that consists of all permutations of the first n natural numbers. sgn is the sign of the permutation.

I don't understand anything... S_n, A_n???

Has sym(n) even order?

Yes it does, for any $n\geq 2$...and it's $Sym_n=:S_n\,,\,\,A_n=$ all the even permutations in $S_n$. If you don't know this then it will probably be a good idea you first read a little about this before you attempt to solve your question.

Tonio
• Jan 16th 2010, 08:58 AM
Amer
Quote:

Originally Posted by Icarus0
Let G be a group. Is it true that there is only one surjective group homomorphism from G to {-1,1}? If this is true how can I proof it?

I need this to proof that f:Sym(n) -> {-1,1} is sgn if f is multiplicative and surjective. Obviously f and sgn are surjective group homomorphisms...

s_n is the group of permutations A_n is the even permutation

$f : S_n \rightarrow \text{ {-1,1} }$

f is multiplicative and surjective want f is sgn

f is multiplicative $f(a.b) = f(a) . f(b)$

$ker(f) = \left\{ \alpha \in S_n : f(\alpha) = identity\; of\; the\; range\; =1\right\}$
• Jan 17th 2010, 08:52 AM
Icarus0
Oh, I see. A_n is the kernel of sgn. We called this Alt(n).
But still I don't understand the followingWhat is the index of a group and what does [S_n:ker f] mean?
• Jan 19th 2010, 09:05 PM
Drexel28
Quote:

Originally Posted by Icarus0
Oh, I see. A_n is the kernel of sgn. We called this Alt(n).
But still I don't understand the followingWhat is the index of a group and what does [S_n:ker f] mean?

I assume, and let NCA validated this, but since $f:S_n\mapsto\mathbb{Z}_2$ is surjective we have that $S_n/\text{ker }f\cong\mathbb{Z}_2$ and in particular that $\left|S_n/\text{ker }f\right|=\left[S_n:\text{ker }f\right]=2$ where $\left[S_n:\text{ker }f\right]$ is the number of cosets formed by $\text{ker }f$ (the index of $\text{ker }f$ in ${S}_n$) therefore since $A_n$ is the only subgroup of order $\frac{n!}{2}$ it follows that $\text{ker }f=A_n$ for every surjection. But, this forces $f\left(S_n-A_n\right)=\{1\}$ which shows that it is the only surjection.