3

Suppose $H\leq G$ and $[G: H]=k$ for some positive integer $k$. How can I show there exists a homomorphism $\varphi:G\longrightarrow S_k$ such that $\textrm{ker}(\varphi)\leq H$?

Notation: Here $S_k$ is the group of permutations of the first $k$ positive integers and $[G: H]$ is the index of $H$ in $G$.

Thanks

PtF
  • 9,655
  • 1
    Maybe consider the action on the cosets of $H$? – angryavian Oct 28 '14 at 23:08
  • Hey, there is theorem that says that every finite group of k elements is isomorphic to a subgroup of $S_k$ can't remember it's name though. Maybe will remember it later... – brick Oct 28 '14 at 23:09
  • I guess it is Caley's theorem, but it says any finite group $G$ of order $n$ is isomorphic to a subgroup of $S_n$, but now we have some more requirements.. – PtF Oct 28 '14 at 23:12
  • @Ptf Yes exactly Caley's theorem. In this case you can compose the projection $\pi : G \rightarrow G/H$ and that specific isomorphism from the theorem and you'll come up with an homomorhism whose kernel is exactly $H$. – brick Oct 28 '14 at 23:17

3 Answers3

2

Hint: Consider the homomorphism which you obtain from the group action $\sigma(x,gH)=xgH$

What is the kernel of this action?

Marm
  • 3,371
  • 1
  • 16
  • 24
1

1. Enumerate the cosets: $$ G/H = \{g_1 H, \ldots, g_k H\}. $$

2. Define the homomorphism $\varphi: G \to S_k$ by the action of $G$ on $G/H$. In more detail, for any $g \in G$ and any $i~(1 \le i \le k)$, $$ g \cdot g_i H = g_j H \quad \text{for some } j~(1\le j \le k). $$ This implicitly defines a permutation $\sigma \in S_k$; i.e. $\sigma(i) = j$. Set $$ \varphi(g) = \sigma. $$

3. Check that this map is a homomorphism. This is one of the axioms of a group action: $$ (g' g) \cdot g_i H = g' \cdot ( g \cdot g_1 H ) $$ In our situation, if $\varphi(g) = \sigma$ and $\varphi(g') = \sigma'$, then $\varphi(g' g) = \sigma' \sigma$.

4. Analyze the kernel. Suppose that $g \in \ker \varphi$. Then, $\varphi(g)$ is the identity permutation in $S_k$, or $$ g \cdot g_i H = g_i H \quad \text{for all } i. $$ This is equivalent to $g \in H$ (consider the identity coset $H$), so $$ \ker \varphi \le H. $$

Sammy Black
  • 25,273
0

Hint: $$\varphi:G\longrightarrow S_k= [f:(xH|x\in G)\rightarrow(xH|x\in G),f\ bijective]$$ $\varphi(g)=f:xH\rightarrow gxH$

mcmat23
  • 1,050