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I found this problem in The Theory of Groups by Marshall Hall.

Let the group $G$ be of order $p^rq^s$. If $G$ has two composition series $1 \unlhd A_1 \unlhd A_2 \unlhd \cdots \unlhd A_r \unlhd A_{r+1} \unlhd \cdots \unlhd A_{r+s} =G$ and $1 \unlhd B_1 \unlhd B_2 \unlhd \cdots \unlhd B_s \unlhd B_{s+1} \unlhd \cdots \unlhd B_{r+s} =G$, where $|A_r| = p^r$ and $|B_s| = q^s$, show that $G$ is a direct product of $A_r$ and $B_s$.

So essentially, $G$ is direct product of its Sylow subgroups, so $G$ is nilpotent. My thought is to show that $A_r$ and $B_s$ are normal in $G$. I can't seem to do so though for some reason. Clearly, $A_r$ is normal in a subgroup of order $p^rq$, which is $A_{r+1}$. Does that imply it's normal in $A_{r+2}$, which has order $p^rq^2$? My thought was to use induction and then show it's normal in $G$, but it doesn't seem to be working or I am missing something?

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Hints: If $p=q$ the result follows.

First think about Jordan-Hölder and what it implies with these two $\textbf{composition}$ series, keep in mind that either $q<p$ or $p<q$.

Without loss of generalization suppose that $p>q$, now think about the last conclusion and what happens between $A_{r}$ and $A_{r+1}$, now think about it for $A_{r+i}$ with $i \in [1,2,...,s]$.

After that think about the size of $A_{r}B_{s}$ and how it relates to $G$. And then ensure the direct product.

And I think it should be it.

Good Luck!

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  • $\begingroup$ Thanks. This hint was enough. I got it. $\endgroup$ – ppham27 Dec 7 '14 at 18:21
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Note that $A_r$ is not only normal in $A_{r+1}$, it is also characteristic (i.e. invariant under all automorphisms of $A_{r+1}$), because it is the necessarily unique largest normal subgroup of $A_{r+1}$ of order a power of $p$. (That has its own notation: $A_r = O_p(A_{r+1})$.)

It follows that $A_r$ is normal in $A_{r+2}$ and it is characteristic in $A_{r+2}$ for the same reason, and hence normal in $A_{r+3}$, etc. So, yes, you can prove it by induction.

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  • $\begingroup$ Thanks for reply. I managed to work out the induction step. $\endgroup$ – ppham27 Dec 7 '14 at 18:22

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