Here is the question, it is from a 2023 PhD qualifying exam at my university:
Let $G$ be a finite group and $H$ a normal subgroup of $G$ of order 5. Prove that if $H$ contains an element not in the center of $G$, then $G$ has an element of order 2.
Here is my solution, but I am curious about the veracity of the highlighted part. It seems like cheating. I also don't see where I used anything specifically from the question for the highlighted part other than the fact that $H$ is of order 5 and hence equivalent to $\mathbb{Z}/5\mathbb{Z}$.
Proof:
Since $G$ is finite, its order will be a multiple of the subgroup, $H$, namely $5 \cdot k$ for $k \in \mathbb{N}, k > 0$. Now, $G$ will have an element of order $2$ exactly when it is an even multiple of $5$. To see this, note that every prime that divides a group's order defines a subgroup in that group. Furthermore, only when the prime's order divides the group's order can it be a subgroup, of course. This has two consequences.
Groups of order $10, 20, 30, \ldots$ have a subgroup of order $2$ (and thus an element of order 2).
Groups of order $5, 15, 25, 35, \ldots$ do not have an element of order $2$. Hence, if our conjecture is true, they must have a normal subgroup of size 5 contained in their center. This is what we'll show.
Every group of order $5 \cdot k$ where $k$ is positive will have a sub-group of order 5. Furthermore, every group of order 5 is isomorphic to $\mathbb{Z}/5\mathbb{Z}$, which means it is Abelian and hence (as a subgroup) it is normal.
Let $x \in H$ and $y \in G$. Then we have: \begin{equation*} \mathbf{xy = yy^{-1} xy = y x y^{-1} y = yx.} \end{equation*} The middle step is because $H$ is Abelian so that once $y^{-1} x y \in H$ (by normalcy), we can commute the factors. This shows that $xy = yx$ in $G$, i.e., that $x$ is in $Z(G)$, as required. \qed
Can we really "commute the factors" as I say? I'm just not sure. Thanks!
Your argument is incorrect.
Note that while $yxy^{-1}$ lies in $H$ and therefore commutes with every element of $H$, you are trying to commute it with $y$, but $y$ is assumed to be in $G$, not in $H$. So the equality $y(y^{-1}xy)=(y^{-1}xy)y$ need not hold.
The question in your title also has a negative answer. The smallest, easiest example is $S_3$, where the normal subgroup of order $3$ is of course abelian, but not central (since $Z(S_3)$ is trivial). Other examples can be found with the subgroups of order $4$ in $D_4$. For an example with a normal subgroup of order $5$, take the dihedral group of order $10$, in which the subgroup of order $5$ generated by the rotations is normal but not central.
In addition, you don't seem to be answering the question! Unless it is your contention that it is impossible for a group to have a normal subgroup of order $5$ unless that subgroup lies in the center.
To actually answer the question, consider the action of $G$ on $H$ by conjugation. This yields a morphism $G\to\mathrm{Aut}(H)$; and since $H$ is cyclic of order $5$, you know that $H$ lies in the kernel of that map, and that $\mathrm{Aut}(H)$ has order $4$. If we are assuming that $H$ has a non-central element, that means that the action is not trivial, so the image of $G$ is not trivial. That means that the image has order $2$ or $4$, which tells you that the order of $G$ is even.