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Suppose a group $G$ where $|G| = 2pq$ where $p$ and $q$ are distinct odd primes. By Sylow theorem there exist subgroups of order $p$ and $q$. Is there a sub-group of order $pq$?

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This is true for abelian groups. Not sure about the general case. – Alex Becker Feb 26 '12 at 11:21
Notice that Geoff's argument below works, essentially unchanged, to establish the much more general fact: if $|G|=2n$, where $n$ is odd, then $G$ has a normal subgroup of index 2. – the_fox Feb 26 '12 at 18:15
up vote 7 down vote accepted

I suppose you intend that $p$ and $q$ are distinct odd primes. In that case, the answer is yes. For let $t$ be an element of order $2$. Then in the action of $G$ on itself by right translation, $t$ acts as a product of $pq$ $2$-cycles. Since $pq$ is odd, $t$ act as an odd permutaton. Hence the elements which act as even permutations in this action form a normal subgroup of index $2$, so of order $pq.$ In general, the answer is no if you allow $p=2$ or $q=2.$ For example, if $p=2$ and $q =3,$ when $G \cong A_4,$ then $G$ has no subgroup of order $pq = 6.$

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Here's a method using the Sylow theorems: note that, assuming $q > p > 2$, we have $n_p \equiv 1 \pmod p$ and $n_q \equiv 1 \pmod q$, where $n_p \,|\; 2q$ and $n_q \,|\; 2p$. If $q \ne 2p-1$, then $n_q = 1$ so the Sylow-$q$ subgroup is normal. If $q = 2p - 1$, then you can argue that $n_p = 1$, so the Sylow-$p$ subgroup is normal. Either way, we see that there is a subgroup of order $pq$, since $PQ$ is a subgroup if one of the subgroups $P$ or $Q$ is normal in $G$.

In the case $p=2$, this does not apply, as Geoff pointed out.

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