Finite Group and normal Subgroup

Question

Let $d,m\in \mathbb{Z}$ with $d,m\geq 1$ and $\gcd(d,m)=1$. Let $G$ be a group of order $dm$ and define the set $X:= \{g\in G | g^d=1\}$.

Show: if $H$ is a normal subgroup of $G$ with order d then $H=X$.

My idea was to use the theorem : if $|H|$ and $(G:H)$ are coprime, then $H$ is the only subgroup of $G$ of order $d$.
Therefore I need to show that $|X|=d$.
I tried this:
For $g\in X$, $g\not= e$, $o(g)/d$ and $o(g) /|X|$ and $o(g)/|G| = dm$.
Since $(d,m)=1$ we have $o(g)\not| m$ and so $|X|/d$.
But now I don't know how to show that $d/|X|$ as well.

Am I right?Or is there a better way to solve this problem?

Answer 1

There are two things you need to prove:

1. $H\subseteq X$, by Lagrange.
2. $X\subseteq H$. To see this, consider the quotient group $G/H$. Suppose $x\in X$. Then $(xH)^d=x^dH=H$. This means that $xH$ has order dividing $d$. Apply Lagrange, we have that $xH=H$, so $x\in H$ as required.

These combine to prove the result.

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Finally - a quick note on your attempt. You said that you want to show that $|X|=d$, and your logic was that you knew that $H$ was the only subgroup of order $d$. However, $X$ is not necessarily a subgroup! For example, the permutations $(123)$ and $(345)$ both have order $3$, but they multiply to give $(1 2 3 4 5)$, a permutation of order $5$.