1
$\begingroup$

I was reading about groups of order 12 and found this:

Let $G$ be a non-abelian group of order $12$. By Cauchy's theorem, it has an element, hence subgroup, $H$, of order $3$

H is not normal in G, then as $[G:H]=4$, there is a homomorphism $ϕ:G→S_4$

I don't understand why there is such homomorphism and what would it be.

I understand that since $[G:H]=4$, I could define the set of the left cosets of $H$ in $G$, $A=\{H, g_1H, g_2H, g_3H\}$ and the group of permutations of $A$ would be isomorphic to $S_4$. But I still don't know how I could define the homomorphism $\phi$. Any help would be greatly appreciated.

$\endgroup$
1
  • 2
    $\begingroup$ The group $G$ permutes its Sylow-$3$-subgroups by conjugation (transitively). So you have a morphism $\phi$, which sens $g\in G$ to the permutation it induces on the four subgroups of order $3$ (by conjugation). $\endgroup$ Commented Feb 18, 2016 at 14:54

2 Answers 2

1
$\begingroup$

A basic fact about group actions is that if a group $G$ acts on a set $X$, then that is equivalent to having a group homomorphism $\phi:G\to\mathrm{Sym}(X)$. This is proved by noting that, for each $g\in G$, the map $$\sigma_g:X\to X$$ given by $\sigma_g(x)=g.x$ is a bijection (with inverse $\sigma_g^{-1}=\sigma_{g^{-1}}$). The assignment $\phi(g)=\sigma_g$ is the desired homomorphism.

Examples:

  1. $G$ acts on itself by left multiplication, so there is a homomorphism $G\to \mathrm{Sym}(G)$ (Cayley's theorem)

  2. The dihedral group $D_{2n}$ acts on the set of vertices of a regular $n$-gon. This yields a homomorphism $D_{2n}\to S_n$

  3. If $H\leq G$ is a subgroup, then $G$ acts on the cosets $G/H$ by left multiplication. This yields a homomorphism $G\to \mathrm{Sym}(G/H)$.

In your question, $[G:H]=4$, so example 3 applies and you get a homomorphism $G\to S_4\cong\mathrm{Sym}(G/H)$.

$\endgroup$
1
$\begingroup$

As pointed out in the other answer, the $G$-action by left multiplication on the left quotient set $G/H$ yields a homomorphism $G\stackrel{\varphi}{\to} S_{G/H}\cong S_4$. But there's more: since $H\not\unlhd G$, then $\tilde gH\tilde g^{-1}\ne H$ for some $\tilde g\in G$. This suffices to get $\ker\varphi=$ $\bigcap_{g\in G}gHg^{-1}$ $=\{1\}$, being $H$ of prime order. Therefore, actually $G$ embeds into $S_4$ or, equivalently, $S_4$ contains an isomorphic copy of $G$.

$\endgroup$
2
  • $\begingroup$ Right, so that's a different action. $\endgroup$
    – i can try
    Commented Mar 28 at 2:40
  • $\begingroup$ And it's faithful. $\endgroup$
    – i can try
    Commented Mar 28 at 2:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .