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When showing isomorphisms, I have sometimes (most likely incorrectly in cases) shown that, for a map $\theta : G \to H$, that $\operatorname{Ker}(\theta) = K \leq H$, means it is $1-1$.

What is the actual link between the two? In order to show that a map is a bijective homomorphism what do I need to show about the kernel?

In the case of an example: if $n \geq 2$, the even permutations of $\{1, 2, \dots, n\}$ form a normal subgroup $A_n$ of $S_n$ of index $2$.

I understand that if you have $\theta : S_n \to S_n/A_n$ by $\theta(g) = \frac{gp(x_1,...x_n)}{p(x_1,...,x_n)}$, that, for $n \geq 2$, $\theta$ is a homomorphism (obvious), it is onto as $\exists$ at least one odd permutation. But instead of showing it is $1-1$ you can show that $\operatorname{Ker}(\theta)$ is $A_n$. Why?

I'm sorry if this question is really obvious.

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    $\begingroup$ A homomorphism of groups $f:G\to H$ is $1-1$ if and only if $\ker f =e_G$. That is, iff $f^{-1}(e_H)=\{e_G\}$. $\endgroup$ – MPW Jan 11 '15 at 14:11
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    $\begingroup$ There is no surjective homomorphism $\theta:S_n \to A_n$ for $n \ge 3$, so your "obvious" assertion must be erroneous! Perhaps you meant $\theta:S_n \to S_n/A_n$? That is onto and has kernel $A_n$. $\endgroup$ – Derek Holt Jan 11 '15 at 14:23
  • $\begingroup$ Sorry yes, changed. $\endgroup$ – mrhappysmile Jan 11 '15 at 14:32
  • $\begingroup$ @MPW - From your answer it makes sense now (I can clearly see that Ker($\theta$) = $A_n$ means it is 1-1). $\endgroup$ – mrhappysmile Jan 11 '15 at 14:45
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For any homomorphism $\theta : G \to H$, the kernel $\ker\theta$ is always a normal subgroup of $G$. The correct statement is that $\theta$ is $1-1$ (injective) if and only if $\ker\theta = \{e_G\}$. Note, this implies nothing about the surjectivity of $\theta$. That is, if $\theta$ is a bijection, then $\ker\theta=\{e_G\}$ but if $\ker\theta=\{e_G\}$, $\theta$ may not be a bijection (it is injective but may or may not be surjective).

As for your example, note that $S_n/A_n$ consists of the cosets of $A_n$ in $S_n$. As $A_n$ has index two, $S_n/A_n$ consists of two cosets: $A_n$ and $\tau A_n$ where $\tau \in S_n$ is an odd permutation. As $A_n$ is a normal subgroup, $S_n/A_n$ is a group (isomorphic to $\mathbb{Z}/2\mathbb{Z}$) with identity element $A_n$. So $\ker\theta = \{\sigma \in S_n \mid \theta(\sigma) = A_n\}$. As $\theta(\sigma) = \sigma A_n$, $\sigma \in \ker\theta \Leftrightarrow \sigma A_n = A_n \Leftrightarrow \sigma \in A_n$, therefore $\ker\theta = A_n$. As $A_n \neq \{e_{S_n}\}$, we see that $\theta$ is not injective.

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