A finite abelian group $G$ has an element $g$ of order $n$, which generates the subgroup $H = \langle g \rangle$. Let $\gamma$ be an element of order $m$ in $G/H$. Show that there exists an element $x \in G$ such that $\lvert x \rvert = \lvert \gamma \rvert = m$.

I copied my instructor's solution, but I missed the last part, and I can't proceed from what's on my notebook.

Suppose that $x_0 \in G$ such that $\overline{x_0} = \gamma$. Since $\lvert \gamma \rvert = m$, $x_0^m \in H = \langle g \rangle$, and we denote $x_0^m = g^k$. Divide $k$ by $m$ to get $k = mq + r$ with $r \in \{0,\dots,m-1\}$. \begin{align} x_0^m &= g^k = g^{mq + r} \\ g^r &= x_0^m g^{-mq} = \left(x_0 g^{-q} \right)^m \end{align} We want to show $r=0$.
Let $x' = x_0 g^{-q}$. Observe that $\overline{\mathstrut x'}=\overline{\mathstrut x_0}=\gamma$ and $m \mid \lvert x' \rvert$ because $$e = x'^{\mathstrut \lvert x' \rvert} = x_0^{\mathstrut \lvert x' \rvert} \underbrace{g^{\mathstrut -q \lvert x' \rvert}}_{\in H}.$$ $$\therefore \lvert (x')^m \rvert = \frac{\lvert x' \rvert}{\gcd(m,\lvert x' \rvert)} = \frac{\lvert x' \rvert}{m}$$

I don't know how to continue with this.

  • $\begingroup$ Hint: Chose your $x_0$ in the beginning such that $k$ is minimal. Then you'll see that $r$ must be zero after line four already. $\endgroup$ – Jesko Hüttenhain Dec 28 '16 at 18:04
  • $\begingroup$ @JeskoHüttenhain So you mean before choosing $x_0$, we should fix $k = \min\limits_{x \in \pi^{-1}(\gamma)} \{\mathrm{ord}_H (x^m)\}$ first? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 28 '16 at 18:49
  • $\begingroup$ That is precisely what I suggest, yes. $\endgroup$ – Jesko Hüttenhain Dec 28 '16 at 18:52
  • $\begingroup$ @JeskoHüttenhain I've made a mistake due to hunger 2 hours ago. It should be $$k = \min\{k' \in \Bbb N^* \mid \exists x \in \pi^{-1}(\gamma) \text{ s.t. } x^m = g^{k'}\}.$$ Then we choose $x_0 \in \pi^{-1} (\gamma)$ such that $x_0^m = g^k$, and we divide $k$ by $m$ so that $k = mq + r$ with $0 \le r < m$. But if $k < m$, then $r = k$, so we can't make $r$ smaller than $k$, and the minimality of $k$ can't be applied to $$g^r = x_0^m g^{-mq} = \left(x_0 g^{-q} \right)^m = \left(x' \right)^m$$ to get a contradiction. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 28 '16 at 22:03
  • $\begingroup$ @GNUSupporter Instead of posting a lot of comments and running from one answer to another, I think would be more useful to edit the question starting by explaining what's the role of $g$ and $n$ in this story. $\endgroup$ – user26857 Dec 29 '16 at 9:31

This argument seems overly complicated (and the conditions overly strict).

Claim. Let $G$ be a finite group and $H\lhd G$ a normal subgroup. Let $\gamma\in H$ be an element of order $m$. Then $G$ has an element $x$ of order $m$.

Proof. Let $\pi\colon G\to G/H$ denote the canonical projection. Pick $x_0\in G$ with $\pi({x_0})=\gamma$. Then $\langle x_0\rangle$ is a cyclic group with $\pi(\langle x_0\rangle)=\langle \gamma\rangle$, hence is a cyclic group of order divisible by $m$. We know that such a cyclic group contains an element of order $m$. $\square$

  • $\begingroup$ yeah, this works. But I think the goal is for $x$ to be in the same coset as $\gamma$ $\endgroup$ – Jorge Fernández Hidalgo Dec 28 '16 at 18:12
  • $\begingroup$ Much cleaner than the instructor's proof in the OP (+1). Minor typo: $\gamma \in G/H$. $\endgroup$ – Bungo Dec 28 '16 at 18:12
  • $\begingroup$ Thanks for your clear explanation! @Bungo Yes, you're right. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 28 '16 at 20:40

Suppose $[x_0]=\gamma$.

It means that $$x_0^m=g^k$$

for some $k$ becuse $x_0^m$ is the unit in the quotient. Now just elevate to the smallest number $t$ making $n|kt$. We get $$(x_0^t)^m = 1$$ thus $ord (x_0^t) \leq m$. Now it can't be strictly less than $m$ because you would find that $g$ has order less than $n$. We were looking for $x:=x_0^t$.

  • $\begingroup$ yeah, this works. But I think the goal is for $x$ to be in the same coset as $\gamma$ $\endgroup$ – Jorge Fernández Hidalgo Dec 28 '16 at 18:12
  • $\begingroup$ I don't understand the second last sentence. If we take $m'=\mathrm{ord}(x_0^t) < m$, then what will happen? $$1=(x_0^t)^{m'}=x_0^{tm'}=x_0^{q'm + r'} = g^{q'k}x_0^{r'}$$ I have to show $x_0^{r'} \in H$? Am I getting the wrong way? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 28 '16 at 21:18
  • $\begingroup$ Simply "apply class" now: $[1]=[g^wx_0^r]=[x_0^r]=[x_0]^r=\gamma^r$ but $\gamma$ has order $m$ and $r<m$ because it comes from euclidean division. $\endgroup$ – Maffred Dec 29 '16 at 2:09

Is this true? Consider $\mathbb Z_4$ and the subgroup $\{0,2\}$. Notice that the odd coset has order $2$ but no element in it has order $2$, they all have order $4$. It is true when $(n,m)=1$ however.

  • $\begingroup$ I think the same, by the way it is not requested that $[x]=\gamma$ if I'm not messing up. $\endgroup$ – Maffred Dec 28 '16 at 18:45
  • 1
    $\begingroup$ Where are you reading the requirement that $x \in \gamma$? $\endgroup$ – Bungo Dec 28 '16 at 18:45
  • $\begingroup$ because of the way the proof is pursued, if we could show that $r=0$ then we would be done, sadly this is not always the case. $\endgroup$ – Jorge Fernández Hidalgo Dec 28 '16 at 18:47
  • $\begingroup$ Thanks. That's the only answer that I've understood now. I'll read the others later. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 28 '16 at 18:48
  • $\begingroup$ the other proofs are correct, I think that the problem is that the teacher wants the element of G to be in the same coset as gamma, which makes the proof harder than jus tfinding an arbitrary element of order m. $\endgroup$ – Jorge Fernández Hidalgo Dec 28 '16 at 18:50

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