$|G|=122 = 2 . 61$
No. of sylow $2$ subgroups $= 1$ or $61 = n_2$
No. of sylow $61$ subgroups $= 1 = n_{61}$
Let the group of order $61$ be $H_{61}$ and the group of order $2$ be $H_2$
Then : clearly both $H_{61}$ and $H_2$ are cyclic.
and if $a ∈ H_2 ⋂ H_{61}$
then, since $|a|$ must divide simultaneously both $2$ and $61$
hence, $H_{61} ⋂ H_2 = \{{e}\}$
Since, $n_{61}=1$ hence, $H_{61}$ is normal and $( H_{61} H_2 )$ is a subgroup $.....(1)$
Now, $| H_{61} H_{2} | = 61 . 2 = 122 ..... (2)$
From $(1)$ and $(2)$ :$ H_{61} H_2 = G ........ (3)$
Since $G = H_2 H_{61}$ and $|G|=122 = |H_2||H_{61}|$, there cannot be any more subgroups of order $2$. Hence $n_2$ must be equal to $1$
Hence, from $(1), (2), (3) : G = H \times K ≅ H ⊕ K ≅ \mathbb Z_2 ⊕ Z_{61} ≅ Z_{122}$
What could have gone wrong in my method? Why am I not getting $D_{61}$ as an answer too?
EDIT : A theorem from Gallian says this :
If a group $G$ is the internal direct product of a finite nymber of subgroups $H_1,H_2, \cdots H_n$, then
$G \approx H_1 \oplus H_2 \oplus \cdots \oplus H_n$
Hence, here $H_2 \approx Z_2$ and $H_{61} \approx Z_{61}$ .
Hence, $G = H \times K \approx H \oplus K \approx Z_2 \oplus Z_{61} \approx Z_{122}$
Where could I have gone wrong?
Thank you for your help.