Proof-verification group order 48 not simple

Question

Is the following proof correct?

Let $G$ be a group of order 48. Let's prove it is not simple.

$\lvert G\rvert=48=2^4\cdot3$. By Sylow's Theorem, $n_3\in\{1,4,16\},\:n_2\in\{1,3\}$.

• $n_3=1$ or $n_2=1$, then $G$ has a normal subgroup (order 3 or 16).

• Suppose $n_2=3$. Then we consider the map: $\Phi:(g,H)\in G\times \text{Syl}_2(G) \longrightarrow gHg^{-1} \in\text{Syl}_2(G)$.

  1. $\Phi(e,H)=eHe^{-1}=H \;\;\;\forall\:H\in \text{Syl}_2(G)$
  2. $\Phi(g_1,\Phi(g_2,H))=\Phi(g_1,g_2Hg_2^{-1})=g_1g_2Hg_2^{-1}g_1^{-1}=\Phi(g_1g_2,H)$

  Then $\Phi$ induces a homomorphism $\varphi:G\longrightarrow S(\text{Syl}_2(G))=S_3$, which is non-trivial. Using the first isomorphism theorem we have $G/\text{Ker}(\varphi) \simeq\varphi(G)=\text{Im}(\varphi) \Longrightarrow \lvert G/\text{Ker}(\varphi)\rvert= \text{Im}(\varphi)$. Using Lagrange Theorem we get $\lvert G\rvert=\lvert\text{Ker}(\varphi)\rvert\dot\lvert\text{Im}(\varphi)\rvert$.

  Now $\lvert\text{Im}(\varphi)\rvert \mid 6=\lvert S_3\rvert$ and $\lvert\text{Im}(\varphi)\rvert \mid 48=\lvert G\rvert$, so $\lvert\text{Im}(\varphi)\rvert\leq 6$.

  We also have $\lvert\text{Ker}(\varphi)\rvert \mid 48$ and $\lvert\text{Ker}(\varphi)\rvert \neq 48$ because the homomorphism is non-trivial.

  Adding both, we have $\lvert\text{Ker}(\varphi)\rvert\geq 8\Longrightarrow \lvert\text{Ker}(\varphi)\rvert\in\{8,12,16,24\}$.

  In any case, for any homomorphism $f$, $\text{Ker}(f)$ is a normal subgroup, so in this case we have $\text{Ker}(\varphi)$ is a normal subgroup of $G$ and its order is neither $1$ nor $48$.

My doubts are the following:

• Is the map $\Phi$ well-defined? Is it a group action?
• Is the map $\varphi$ non-trivial because the definition of $\Phi$?
• What would happen if $\lvert\text{Ker}(\varphi)\rvert=1$?

Answer 1

First, I assume you mean $$\Phi:(g,H)\in G\times \text{Syl}_2(G) \longrightarrow gHg^{-1} \in\text{Syl}_2(G)$$

• It is indeed well defined because $p$-Sylow subgroups form conjugacy classes.

• That the map $\varphi$ is non-trivial is seen from how it is defined: $$\varphi:G\to\mathrm{Perm}(\text{Syl}_2(G))\simeq S_3 \atop g\mapsto m_g := \Phi(g,\cdot)$$ If it were trivial, it would mean that conjugation by any $g$ has no effect on $2$-Sylows (why?) and this cannot be, since $n_2 = 3$.

• If $\ker\varphi = \{1\}$ you should still be able to conclude from $\lvert G\rvert=\lvert\text{Ker}(\varphi)\rvert\dot\lvert\text{Im}(\varphi)\rvert$