How many non-isomorphic groups of order 122 are there? How many non-isomorphic groups of order 122 are there?
Let $G$ be  a group of order 122.No of Sylow 61 subgroups of order 61=1 and hence it is normal say it is $H$.
No. of Sylow 2 subgroups of order 2 is either 1 or 61.If it is 1 then $G\cong \mathbb Z_2\times \mathbb Z_{61}$
I have two questions frm now on:
Since we have two different choices for number of Sylow 2 subgroups can we conclude from here that there exist two non-isomorphic groups of order 122? Please help
My problem is when it is 61.Then we have 61 Sylow 2 subgroups of order 2.In this case how should I find $G$?Any help
 A: It can be shown that if the order of a group is $2p$ for some odd prime $p$, then either 
$G \cong Z_{2p}$ or $G \cong D_p$
By the first Sylow Theorem, $G$ has subgroups $H, K$ such that:
$$\newcommand{\angle}[1]{\langle #1\rangle} H = \angle{a},\,K = \angle{b},\,|a| = p,\,|b| = 2$$
Since [G:H] = 2, H is a normal subgroup of G; this means
$$\newcommand{\set}[1]{\left\{ #1\right\}}HK = \set{hk \mid h \in H, k \in K} < G$$
Since every element of $H$ has order 1 or $p$, and every element of $K$ has order $1$ or $2$,
$$H \cap K = \set{g \in G \mid |g| = 1} = \set{e}$$
This gives us
$$|HK| = \frac{|H|\cdot|K|}{|H \cap K|} = 2p/1 = 2p = |G|,$$ so $HK = G$
$$G = \set{a^i b^j \mid |a| = p, |b| = 2, ba = ?}$$
It remains to be seen what $ba$ equals; that will give us the group table and the group.
Since $H$ is normal in $G$, $bab^{-1}\in H$, $bab^{-1} = a^i$ for some integer $i$
$$bab^{-1} = a^i, a = b(a^i)b^{-1}, a = b [b(a^i)b^{-1}]^i b^{-1}, a = a^{i^2}, a^{i^2 - 1} = e$$
This means that $|a| = p\mid i^2-1$, so $i =p+1\equiv 1\pmod{p}$ or $i =p -1$
$$i = 1,\quad ba = ab,\quad G\text{ Abelian}\implies G \cong \Bbb{Z}/2\Bbb{Z} \times \Bbb{Z}/p\Bbb{Z} \cong \Bbb{Z}/2p\Bbb{Z}$$
$$i = p-1,\quad ba = a^{p-1}b,\quad ba = a^{-1}b \implies G \cong D_p$$
Substitute 61 for $p$ and you're good to go
