Prove that every group of order $49$ contains a subgroup of order $7$. Prove that every group of order $49$ contains a subgroup of order $7$.
We know that a group of prime order is always cyclic. Also, we know that if the order of a group is $pq$, where $p,q$ are prime integers, then a subgroup $H\neq G$ is a cyclic subgroup. But here in this question, we don’t know whether such a non-trivial  subgroup $H$ of $G$ exists. If it would have existed, then we could have said, by Lagrange’s theorem, that the subgroup has an order of $7$ and hence $H$ is cyclic. But such an information is not given. So how do we prove it? I am not quite getting it . . . Also, there might be posts concerning the similar topic on this site but I can’t seem to find it either . . .
 A: Every nontrivial element of $G$ has order $7$ or $49$ (Lagrange). In the former case, you are done. In the second case, say $x$ an element of order $49$; then, the element $x^7$ has order $7$.
A: Hint
Cauchy's theorem ensures an element of order $7$.
Since you don't know Cauchy, we do the following:  Either the group is cyclic,  in which case it contains a subgroup of every order dividing the order of the group;  or it contains an element of order $7$. (Of course we used Lagrange.)

Alternative: every group of order $p^2$ is abelian.   Thus by the structure theorem,  either we have $\Bbb Z_{49}$ or $\Bbb Z_7×\Bbb Z_7$.
A: $G$ is a finite Group of order $49$. Since $7$ is  a prime so $G$ is a $7$-group and so $|Z(G)|>1$.
As $G$ is $7$-group so $Z(G)$ is also a $7$-group.
So, $7|Z(G)$ so there exists an element $a$ belongs to $Z(G)$ such that $o(a)=7$.
Since $o(a)=7$ and $7$ is prime we will get a cyclic subgroup $H$ generated by $a$. Hence we get a subgroup of order $7$.
If we look into it more closely we will observe that the subgroup is also normal.
