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I am in the middle of reading a proof.

So far I have a group $G$ of order $q^2p^2$, both prime. I have two Sylow $q$-subgroups $Q_1$ and $Q_2$ both of order $q^2$. I know $Q_1\cap Q_2\neq \{e\}$, $|\langle Q_1, Q_2\rangle| = q^2p$. So, $[G:\langle Q_1, Q_2 \rangle ] = p$.

Apparently, I immediately know I have a homomorphism $\varphi: G \rightarrow S_p$? And its kernel is nontrivial?

The rest of the details are left out and I am having lots of trouble understanding how we know that's the case (and it's so easy that it's just completely left out). Thank you so much!

Shaun
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MCMath
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1 Answers1

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Given any subgroup $H$ of $G$ with $[G:H]=k$, you can let $G$ act on the set of left cosets of $H$ in $G$ by left multiplication, $$g\cdot (xH) = (gx)H.$$ This is an action on a set with $k$ elements.

An action of $G$ on a set $X$ is equivalent to a group homomorphism $G\to\mathrm{Perm}(X)$, the group of permutations of $X$; and if $|X|=k$, then $\mathrm{Perm}(X) \cong S_k$. Thus, if you have a subgroup $H$ of $G$ of index $k$, then this lets you define a group homomorphism $G\to S_k$ by having $G$ act on the left cosets of $H$.

In this case, why do we know the kernel is nontrivial? Because the largest power of $p$ that divides $|S_p|=p!$ is $p$. So the image of $G$ has order not divisible by $p^2$. Since $|G|$ is divisible by $p^2$, the kernel must have order at least $p$ and hence be nontrivial.

Arturo Magidin
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