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Although it is a duplicate of If $H \triangleleft G$ and $|G/H|=m$, show that $x^m \in H$ for $\forall x \in G$, I didn't find the answer all too useful as some parts are just said 'to be true' without much explanation. Also, from the answer I couldn't conclude the necessity of $H\triangleleft G$, which was required in the following question:

Suppose $H\triangleleft G$ and $n=[G:H]<\infty$. Prove $\forall_{g\in > G}:g^n\in H$. Also give an explicit example from which it can be concluded that it is necessary that $H$ is a normal subgroup of $G$.

I approached it relatively similarly to the answer from the duplicate question;

$[G:H]=\#(G/H)=n$. As all orders of elements of a group are divisors of the order of the group, in particular it holds that $(gH)^n=eH$ for all $g\in G$.

After all: suppose $\operatorname{ord}(gH)=m$ such that $n=pm$ for a $p\in\mathbb{N}$. By definition: $$(gH)^m=eH$$$$(gH)^n=(gH)^{mp}=((gH)^m)^p=(eH)^p=eH=H$$

As $(gH)^n=g^nH=H$, we have that $g^n\in H$.

I don't particularly see where I used that $H$ must be normal in $G$. Is it implicit in the assumption that $G/H$ exists as a group, of which we use the order?

Marc
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  • See also https://math.stackexchange.com/questions/573050/if-h-is-a-subgroup-of-g-of-finite-index-n-then-under-what-condition-gn – lhf Jun 14 '18 at 13:54

2 Answers2

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You used that $H\lhd G$ in order to have that $G/H$ is a group.

If we drop the condition that the subgroup $H$ is a normal in $G$, the conclusion may no longer hold: Consider $G=S_3$ and $H=\{e,(1\,2)\}$, a subgroup of index $3$. Then for $g=(1\,3)$, we have $g^3=g\notin H$.

Remark on the problem statement: Strictly speaking, a single example where the premise and the conclusion are false ($\exists x\colon \neg P(x)\land\neg C(x)$) does not yet mean that the premise is necessary for the conclusion ($\forall x\colon C(x)\to P(x)$).

Hagen von Eitzen
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Yes, whenever you write something like $(gH)^m$, you are implicitly using normality of $H$ in order for this to be well defined. For a concrete counterexample, take for instance the cyclic subgroup $\{1,(1,2,3),(1,3,2)\}$ inside $S_5$. It has index $40$ inside $S_5$, but if you take, for instance, $(3,4,5)$, you will have $(3,4,5)^{40}=(3,4,5)\not\in H$.

TomGrubb
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