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G is a group and H is a normal subgroup of G.

Let $m$ be a fixed integer. If $x^m \in H$ for every $x \in G$, then the order of every element of $G/H$ is a divisor of m.

I am getting quite confused by this question and I attribute it to my very thin understanding of quotient groups (since we just started learning about them 2 days ago).

But here is what I understand:

I must prove that an element $Hx \in G/H$ has order $n$ such that $n|m$.

But I don't understand how knowing that $x^m \in H$ helps me. I think I understand how to prove the converse but the forward direction confuses me.

JxxYsde3
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1 Answers1

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Use the fact that $(Hx)^n=Hx^n=H$ since $H$ is normal, this implies that the order of the cyclic group generated by $Hx$ divides $m$.

Tsemo Aristide
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  • I'm sorry, I don't think I follow ... perhaps I need to study a little bit more before attempting to answer questions. But how does $(Hx)^n = H$ imply that m divides n? Isn't m just what x is raised to? – JxxYsde3 Oct 26 '17 at 15:52
  • Let $p:G\rightarrow G/H$, $(Hx)^n=H$ is equivalent to say that $(Hx)^n$ is the neutral of $G/H$. Divide $n$ by $m, n=am+r, r<m$, $(Hx)^n=(Hx)^{am+r}=(Hx)^{am}(Hx)^r=(Hx)^r$ since $m$ is the smallest integer different of zero such that $(Hx)^m=H$, you deduce that $r=0$. – Tsemo Aristide Oct 26 '17 at 15:57
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    So would this solution be correct: $(Hx)^n$ = $H$ which implies that $x^n \in H$. Since $x^m \in H$ for all $x \in G$ this implies that m|n – JxxYsde3 Oct 26 '17 at 17:57