Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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, $NM = G$.

share|improve this question
add comment

3 Answers

up vote 2 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 $pq$.

share|improve this answer
    
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
add comment

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 $pq$.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.