Prove that every group $G$ of order $98,$ having just one element $g$ of order $2$ is an abelian group. Prove that every group $G$ of order $98,$ having just one element $g$ of order $2$ is an abelian group.
I was thinking about use that the group $G = 7^2·2$ so I have a group $N_2$ and $H_7$ and I need to prove what are the subgroups, watch if i can put G like cyclic groups one $Z_2$ and the other ones with no elements order 2. So I have $G\simeq Z_2$ x $Z_n$ x ... I hope it is enough.
 A: You have the right idea in terms of showing that $G$ is a direct product of abelian groups, but you are missing some details. Moreover, the subgroup of order $7^2$ need not be cyclic (not sure if you are actually trying to say this tbh).
By the first Sylow theorem we have a subgroup $H_7$ of order $7^2$ and a subgroup $N_2$ of order two $2$, which you also seemed to have figured out. The latter is unique by the assumption that there is only one element of order $2$. We will now show that also the first one is unique:
We know by the third Sylow theorem that the number of Sylow-$7$ subgroups needs to divide $2$ and that this number is congruent to $1 \text{ mod } 7$. Hence it must be odd and is therefore $1$. This tells us, that both of these groups are normal subgroups of $G$.
Since the intersection of $H_7$ and $N_2$ is trivial by Lagrange, we know that $H_7 \cdot N_2$ must be $G$ by comparison of the cardinalities. To summarize, we have
$\bullet$ $H_7$ and $N_2$ are normal subgroups of $G$
$\bullet$ $H_7 \cap N_2 = 1$
$\bullet$ $H_7 \cdot N_2 = G$
which means that $G \cong H_7 \times N_2$. Now note, that a group of prime order is cyclic and hence abelian and that a group of order $p^2$ for some prime $p$ is either cyclic or a direct product of two cyclic groups and hence also abelian. Thus $H_7 \times N_2 \cong G$ is abelian, as wanted.
