# About minimal subgroup of a finite group.

Let $G$ be a finite group such that $G=PQ$ where $P$ is a Sylow $p$-subgroup of $G$ and $Q$ is a normal Sylow $q$-subgroup of $G$. I need to prove the following:

(1) If $H/Q$ is a subgroup of $PQ/Q$ of order $p$, then $H/Q=\langle x\rangle Q/Q$ where $x \in P$ and the order of $x$ is $p$.

(2) Suppose that $p=2$. If $H/Q$ is a cyclic subgroup of $PQ/Q$ of order $4$, then $H/Q=\langle x\rangle Q/Q$ where $x \in P$ and the order of $x$ is $4$.

I think that $H/Q=\langle xQ\rangle$ which is isomorphic to $\langle x \rangle / (\langle x \rangle \cap Q)$, but how to prove that $\langle x \rangle \cap Q=\{1\}$.

Thanks.

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Use the isomorphism theorems:

$$(1)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\langle x\rangle Q/Q\cong \langle x\rangle/\left(\langle x\rangle\cap Q\right)\cong\langle x\rangle$$

since assuming $\,p\neq q\,$ , we get that $\,x\in P\Longrightarrow \langle x\rangle\cap Q=\{1\}$

To end this, just observe that $\,H/Q\,$ is a subgroup of order $\,p\,$ then $\,H/Q=\langle hQ\rangle=\langle h\rangle Q\,$ , with $\,\mathcal Ord(x)=p\,$ , and since any $\,p-$subgroup of $\,G\,$ is contained in some Sylow $\,p-$subgroup...

Now point (2) must follow at once from (1) above, introducing some little changes...

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Sorry, I do not understand how $\langle x \rangle \cap Q=\{1\}$. – user28083 Feb 5 '13 at 15:55
@user28083: $x\in P$ so $|x|=p^n$, for some positive integer $n$. If $x$ wants to be in $Q$ as well, so $|x|\big||Q|$ and so $|x|=q^m$, for some positive $m$. Since $p,q$ are distinct primes so $m=n=id$. – Babak S. Feb 5 '13 at 18:12
+1 Well explained, @BabakSorouh. Of course, under the assumption that $\,p\neq q\,$ , as noted. – DonAntonio Feb 5 '13 at 18:46
@BabakSorouh, you say that $h$ is contained in some Sylow $p$-subgroup, but how do you know it is in $P$, as requested? – Andreas Caranti Feb 5 '13 at 19:12

For (1), you may argue like this. Clearly we have $P H = G$, as already $P Q = G$. A general result tells you that for subgroups $A, B$ of a finite group $G$ we have $\lvert B A : A \rvert = \lvert B : B \cap A \rvert$. ($B A$ need not be a subgroup, by $\lvert B A : A \rvert$ we just mean the number of cosets $b A$, for $b \in B$.)

In our case this reads $$\lvert G : H \rvert = \lvert P H : H \rvert = \lvert P : P \cap H \rvert.$$ Now note that $H$ has order $p \cdot \lvert Q \rvert$, so $\lvert G : H \rvert = \lvert PQ : H \rvert = \lvert P \rvert / p$, and it follows that $P \cap H$ is a subgroup of order $p$ of $P$. This is your $\langle x \rangle$.

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By Dedekind modular law, $H=PQ \cap H =(P \cap H)Q$. Since $P \cap Q=\{1\}$, then by the second isomorphism theorem, we have $P \cap H \cong (P \cap H)/ \{1\}=(P \cap H)/((P \cap H) \cap Q) \cong (P \cap H)Q/Q= H/Q$.
Now, for (1), if $H/Q$ is of order $p$, then $P \cap H$ is of order $p$.
Similarly, for (2), if $H/Q$ is cyclic of order $4$, then $P \cap H$ is cyclic of order $4$.
In both cases, we can take $x$ to be the generator of $P \cap H$ and so $H/Q=\langle x \rangle Q/Q$.