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Let $G$ be a finite group and let $H$ be a normal subgroup of $G$. Prove that the order of the element $gH$ in $G/H$ must divide the order of $g$ in $G$.

I see that if we have: $H = \{e, h_1, ..., h_2\}, gH = \{g, gh_1, ..., gh_2\}, ..., g^{k-1}H = \{g^{k-1}, g^{k-1}h_1, ..., g^{k-1}h_2\}$, then the only true deciding element in $H$ in $e$, because no matter what $g$ you use you will produce the original coset once $g^{|g|} = e$.

But this seems to show that $|g| = |gH|$.

Am I missing something here? Because this does divide it, but only trivially.

Oliver G
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  • The title of the question and the actual question doesn't quite match. In the title we are asked to prove whether $|gH|∣|g|$ but in the question OP asks whether he/she is missing something, i.e., whether his/her proof is correct. –  Aug 20 '16 at 01:19

3 Answers3

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By the division algorithm, we can write, $|g| = q\cdot |gH|+r$ for integers $q,r$ where $0 \leq r < |gH|$. Notice that $g^{|g|} = e$, so

$$(gH)^{|g|} = (g^{|g|})H = eH = H$$

and

$$(gH)^{|g|} = (gH)^{q\cdot|gH|+r} = \left((gH)^{|gH|}\right)^q \cdot (gH)^r = H^q \cdot (gH)^r = (gH)^r,$$

thus,

$$(gH)^r = H.$$

Since $H$ is the identity element of $G/H$, then the above implies that if $r$ is nonzero, then $r$ is at least the order of $gH$; however, from before we have that $r < |gH|$, thus the only possible value of $r$ is $0$. Hence, $|g| = q \cdot |gH|$, thus $|gH|$ divides $|g|$.

benguin
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Writing $g^{\lvert g\rvert}H=H$ only means $\lvert g\rvert$ is a multiple of the order of $gH$.

Indeed we very well may have $g^rH=H$ for smaller values of $r$, since $g^rH=H$ means $g^r\in H$, not $g^r=e$.

As a counterexample, consider the cyclic group $C_4$, $g $ a generator, $H$ the subgroup $\langle\mkern 1mu g^2\rangle$. $G/H$ is the cyclic group $C_2$, generated by $gH$. So $\lvert gH\rvert=2$, while $\lvert g\rvert=4$.

Bernard
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  • But how exactly would I show this? I see that $(gH)^{|g|} = g^{|g|}H = eH = H$, and the book I'm reading suggests to use the fact that "$a^k = e \Rightarrow |a|$ divides $k$" but I don't see how this would relate. – Oliver G Aug 19 '16 at 13:43
  • From $(gH)^{\lvert g\rvert}=H$ it's straightforward to deduce $\lvert gH\rvert$ divides $\lvert g\rvert$. I don't see what the problem is? – Bernard Aug 19 '16 at 18:23
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The equality is not true in general because if $g\in H\setminus\{e\}$ then $gH=H$ but the order of $g$ is not $1$.

To take a concrete example, let $(G,\circ)$ be $(\mathbb{Z}_2 \times \mathbb{Z_2},+)$ where the "addition" is done component wise. Let $H:=\{(0,0),(1,1)\}$. Now observe that, $$(1,1)+H=H$$ and hence $\lvert (1,1)+ H\rvert=1$ but $|(1,1)|=2$.

  • Is this a counter example? The order of $(1,1) H$ in $G/H$ divides the order of $(1,1)$ in $G$, as the OP claims. – peter a g Aug 19 '16 at 13:47
  • @peterag:This is not a counterexample to the claim that OP wants to prove (because it is true). But it is a counter example to the equality OP has written in the post (admittedly, the title of the question and the actual question doesn't match). –  Aug 19 '16 at 13:58