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This is question 2.9 #9 from Topics in Algebra by Herstein:

If $o(G)$ is $pq$ where $p$ and $q$ are distinct prime numbers and if $G$ has a normal subgroup of order $p$ and a normal subgroup of order $q$, prove that $G$ is cyclic.

Let $N$ denote the normal subgroup of order $p$ and $M$ denote the normal subgroup of order $q$. Here are a couple things I noted while exploring the problem:

  • $N$ and $M$ are cyclic, since they have prime order.
  • $G = NM$ [1].

Any hints where to look next?

[1] Since $N,M$ are normal, $NM \le G$. Then $o(NM) \mid o(G)$, so $o(NM)$ is either $1$, $p$, $q$, or $pq$. The first three don't work because they force one of $N$ or $M$ to be $(e)$, so we have $o(NM) = pq$. Therefore, $NM = G$.

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up vote 5 down vote accepted

Look first at Lagrange's theorem, that tells you that $N \cap M = \{ e \}$.

This implies that if $a \in N$, $b \in M$, then $a b = b a$, because $$ (ba)^{-1} a b = a^{-1} b^{-1} a b = (a^{-1} b^{-1} a) b \in M, \qquad a^{-1} b^{-1} a b = a^{-1} (b^{-1} a b) \in N, $$ since $N, M \triangleleft G$,

If you take in particular $a \ne 1$ (so that $a$ has order $p$) and $b \ne 1$ (so that $b$ has order $q$), then $ab$ will have order $pq$.

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That was trickier than I expected. I followed along, but I doubt I would have figured it out myself. I'll have to come back to this problem in a few days to make sure I can still do it. – Rainbow Apr 9 '13 at 8:02
@Rainbow You don't have to deal with elements in this problem, see my answer below. – user38268 Apr 9 '13 at 8:05

Since $M$ is normal we can form $G/M$. Now $G/M$ has order $p$ and thus is cyclic, in particular abelian. It follows that $[G,G] \subseteq M$. Now similarly we find also that $[G,G] \subseteq N$ and so $$[G,G] \subseteq M \cap N = \{e\}.$$ Now recall that $[G,G]$ is such that $G/[G,G]$ is abelian. Since $[G,G] =\{e\}$ we conclude that

$$G/[G,G] \cong G/\{e\} \cong G$$

is abelian too. The structure theorem for finite abelian groups now gives that $G \cong \Bbb{Z}/p\Bbb{Z} \times \Bbb{Z}/q\Bbb{Z}$ or $G \cong \Bbb{Z}/pq\Bbb{Z}$. However the former is isomorphic to $\Bbb{Z}/pq\Bbb{Z}$ by CRT and so in any case we see that $G$ is cyclic of order $pq$.

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You'll want to show everything commutes. This can be wrapped up succinctly via $[M,N]=1$ using the commutator bracket, since if everything in $M\cup N$ commutes then everything in the whole group $G=\langle M\cup N\rangle$ commutes. Note the identity $[A,B]\le A\Leftarrow B\le N_G(A)$ for this purpose.

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Although my proof has same idea as suggested by @Andreas Caranti.

But let me just highlight it this viewpoint.

Let H be subgroup of order p and K be subgroup of order q. Then $H \cap K = \emptyset$.

And $hk = kh\quad \forall h \in H$ and $\forall k \in K$ because H and K normal as proved by Andreas.

Hence G isomorphic to $H\times K$(easy to find isomorphism now). Then since H isomorphicc to $Z_p$ and K isomorphic to $Z_q$. Hence G isomorrphic to $Z_p \times Z_q$ which is then siomorphic to $Z_{pq}$ which is cyclic.

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