# Proof-verification group order 48 not simple

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 $\lvert\text{Ker}(\varphi)\rvert=1$?
• is the direct product external? – Jorge Fernández Hidalgo Jan 6 '15 at 0:11
• @JorgeFernández $G$ is a group and $\text{Syl}_2(G)$ is a set, so I think that is external – user203327 Jan 6 '15 at 0:17

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 $\lvert G\rvert=\lvert\text{Ker}(\varphi)\rvert\dot\lvert\text{Im}(\varphi)\rvert$
• Yes, I meant that. I don't get part "conjugation by any $g$ has no effect on $2$-Sylows". Then, if $\text{ker}(\varphi)=\{1\}$ then the image would have to have order 48 and it is not possible? – user203327 Jan 6 '15 at 10:54
• @user203327 That is correct. For the part you mentioned, it's like this: The set $\mathrm{Perm}(\text{Syl}_2(G))$ is the group of bijections from $\text{Syl}_2(G)$ to itself. If all $g$ are mapped to $\Phi(g,\cdot)$ (conjugation by $g$), and this is always the trivial bijection aka $\mathrm{id}$, this means that conjugation by $g$ leaves $2$-Sylows untouched. – GPerez Jan 6 '15 at 12:35
• And if $g$ leaves $2$-Sylows untouched then $gHg^1=H$, so it would be normal – user203327 Jan 6 '15 at 12:38
• @user203327 That too. Unfortunately that only lets you say: "and if this is the case, we are finished, so now suppose that $\varphi$ is trivial... and you wouldn't be able to deal with the case $\mid\ker\varphi\mid = 48$. You need to note that you've already excluded the case $n_2 = 1$, and since all $2$-Sylows are conjugate, it's impossible that all conjugations leave $2$-Sylows untouched. – GPerez Jan 6 '15 at 12:54