Let $G$ be group of order $122 = 61 \cdot 2 = p \cdot q$ , where $p < q$ are primes. I know that there exists a unique non abelian group of order $pq$ and one abelian non isomorphic group of order $pq$. I think here are two non isomorphic of order $122$.

Please check my answer, it is right or not.

  • 1
    $\begingroup$ There are indeed precisely two groups of order $122$ up to isomorphism. Your reasoning is somewhat incoherent, though. $\endgroup$
    – Servaes
    Jun 24, 2014 at 8:48

3 Answers 3


The general classifiaction of groups of order $pq$ ($p < q$ prime) is the following:

There is always the cyclic group $\mathbb Z/pq\mathbb Z$.

For $q\not\equiv 1$ mod $p$, this is the only isomorphism type of groups of order $pq$.

For $q \equiv 1$ mod $p$, there is a single additional isomorphism type, which has the form $\mathbb Z/q\mathbb Z\rtimes \mathbb Z/p\mathbb Z$. If $p = 2$ (and $q$ is odd), a suitable representative of this second isomorphism type is the Dihedral group $D_q$.

So in your case of order $122$, there is $$\mathbb Z/122\mathbb Z \qquad \text{and} \qquad D_{61}.$$


By Cauchy, $G$ has a (cyclic) subgroup $N$ of order $61$, which has index $2$ and hence is normal. Also $G$ has an element of order 2, whence a subgroup $H$ of order $2$, $G=HN$ and $H \cap N=1$. So $G$ is either a direct or semi-direct product of $H$ and $N$.


There are only two non isomorphic groups of order $2p$ where $p$ is a prime number greater than $2$. One is isomorphic to $D_{2p}$ and other one is Cyclic.

Looking for a simple proof that groups of order $2p$ are up to isomorphism $\mathbb{Z}_{2p}$ and $D_{p}$ .


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