(related to this question: Finite Group and normal Subgroup)
Let $d,m\in \mathbb{Z}$ with $d,m\geq 1$ and $\gcd(d,m)=1$. Let $G$ be a group of order $dm$ We define the set $X:= \{g\in G | g^d=1\}$ and $H$ a subgroup of $G$.
If $H\subseteq X$ then $|H|$ divides $d$.
I tried to show it the other way around: If $\#H\not | d$ then $H\not\subseteq X$.
For every element $h\in H$ we have $ord(h)| \# H$. Let $d:= q_1*...*q_k$ and $\# H= p_1, ..., p_r$ with all $p_i\not=q_j$ since $\#H\not| d$. Then we have that $ord(h)=p_{i_1}*...*p_{i_l}$ for some of the $p_i$ in the product of $\# H$. But $h^d=1$ iff $ord(h)|d$. But since all $p_{i_\alpha}\not= q_j$ we have $h^d\not=1$ for all $h\in H$.
I feel that I am missing something. Does this way make sense?
Best, Luca
I don't think all $p_i \neq q_j$ necessarily follows from $|H| \nmid d$. That would require $\gcd(|H|,d) = 1$ which is a stronger condition.
I'm guessing $H$ has to be a subgroup. Then note that $|H| \mid |G|$ and for $h \in H$, $h^d = e$. Suppose that $p$ is a prime which divides $|H|$. By Cauchy's theorem, there exists an element of order $p$ in $H$, so $p \mid d$ and $p \nmid m$. Therefore, we have that $\gcd(|H|,m) = 1$. Furthermore, $|H| \mid |G| = dm$ so $|H| \mid d$.