I was working with A.A. Albert's Fundamental Concepts of Higher Algebra and noticed I was able to complete one of the exercises without actually using all the assumptions. I imagine there must be something wrong with my proof but can't figure out exactly what it is. The exercise is as follows:
"Let $ G $ be a cyclic group of order $ n=mq $ where $ m > 1 $, $ q > 1 $, and $ m $ is prime to $ q $. Show that the subset $ H $ of all elements of $ G $ whose orders divide $ m $ is a subgroup of $ G $."
My attempt doesn't use the assumption that $ m $ is prime to $ q $, which I take to mean that $m$ and $q$ are relatively prime. Anyway, I make use of some pretty basic lemmas, namely, that if $ g^{m}=e $ then the order of $ g $, which I denote $o(g)$, must divide $m$. I also use the fact that $o(g)=o(g^{-1})$. I'll skip the proof of those. Anyway, I simply have to show that if $ h_{1},h_{2} \in H $ then $ h_{1}h_{2}^{-1} \in H $. So, since $ G $ is cyclic we can write $ h_{1}=g^{k_{1}} $ and $ h_{2}^{-1}=g^{-k_{2}} $. Now we consider
\begin{align*} (g^{k_{1}}g^{-k_{2}})^{m} &= (g^{k_{1}-k_{2}})^{m} \\ &= g^{m(k_{1}-k_{2})} \\ &= g^{mk_{1}}g^{-mk_{2}} \\ \end{align*} Now since $ h_{1},h_{2} \in H $ we have that $o(g^{k_{1}}) = d_{1}$, $o(g^{k_{2}})=d_{2} $, and $ d_{1}q_{1} = m = d_{2}q_{2} $ which gives us
\begin{align*} &= g^{d_{1}q_{1}k_{1}}g^{-d_{2}q_{2}k_{2}} \\ &= (g^{d_{1}k_{1}})^{q_{1}}(g^{-d_{2}k_{2}})^{q_{2}} \\ &= e^{q_{1}}e^{q_{2}} \\ &= e \end{align*} Thus $ o(h_{1}h_{2}^{-1}) \vert m $ which implies that $ h_{1}h_{2}^{-1} \in H $. Therefore $ H $ is a subgroup.
Again, I'm not sure where I went wrong. Pointing out the flaw without revealing the correct proof would be much appreciated. Thank you!
The assumption isn't needed. In fact, in any commutative group $G$ the set $H$ of elements with order dividing $m$ is a subgroup. That's because the condition for $x$ to belong to $H$ is equivalent to saying that $x^m=e$. So $H$ is the kernel of the morphism $x\mapsto x^m$ from $G$ to $G$.