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!
Best Answer
$H$ is normal, so it can be written as the set-theoretical disjoint union of $G$-conjugacy classes. One of these is $\{1\}$ and we are left with $4$ elements. Since $H \cap Z(G)=1$ no further singletons can arise, so either the cardinalities of the other $ G$-conjugacy classes are $2+2$ or $4$. Either way, $2$ divides the cardinality and on its turn this number divides $|G|$. Hence $|G|$ must be even and we finish with an appeal to Cauchy's Theorem.