Consider the theorem

Let $G$ be a finite Abelian group with order $|G|=p^n$ and $a$ an element of maximal order in $G$, then there is a subgroup $H$ of $G$ such that $G\cong |a|\times H$.

I'm interested only in the begin of the proof (I've seen some online and all go a similar way through). Consider $H$ the maximal subgroup of $G$ with $H\cap \langle a\rangle =\{e\}$.

No one says in his proof why this subgroup exist. I must miss something very elementary. I undestand the following: Lets take $b\in G-\langle a\rangle$, then $\langle a\rangle\cap \langle b\rangle=\{e\}$. Now the problem arises from the maximality. Let's take $c\in G-\langle a\rangle\langle b\rangle$ (we consider that $b$ and $c$ exist otherwise we are already done).

What can I say about $\langle b\rangle\langle c\rangle$? Could we build $H$ iteratively $H=\langle b\rangle\langle c\rangle\cdots$?

  • $\begingroup$ If $H$ is a subgroup satisfying the condition that $\langle a \rangle \cap H = \{e\}$, then any subgroup of $H$ also satisfies this condition. The minimal subgroup satisfying this condition is $\{e\}$. In fact, since this is one subgroup satisfying the condition, and since the group is finite (so there is no infinitely ascending chain of subgroups), we know that "maximal subgroup $H$ such that $H \cap \langle a \rangle = \{e\}$" is well defined and must exist. $\endgroup$ – M. Vinay Mar 29 at 12:18
  • $\begingroup$ Your question about $\langle b \rangle$ and $\langle c \rangle$ is not really clear (to me at least). Are you saying that what you have constructed seems to contradict maximality or the possibility of there being such a maximal subgroup? $\endgroup$ – M. Vinay Mar 29 at 12:19
  • $\begingroup$ @M.Vinay Your first point. You could have $H$ and $H'$ satisfying the condition but neither $H\le H'$ or $H'\le H$. I mean lattice of subgroups. $\endgroup$ – Aaron Lenz Mar 29 at 12:26
  • $\begingroup$ Oh, so you're worried about uniqueness of the maximal subgroup. But that's not required. You can have two different subgroups maximally satisfying the property. Unless you want to additionally prove that there's a unique one. $\endgroup$ – M. Vinay Mar 29 at 12:27
  • $\begingroup$ @M.Vinay Your second point. with $b$ and $c$, I'm trying to build $H$. My conjecture is that with iterative finding of excluded elements we build $H$ by subgroup multiplication of the spanns. $\endgroup$ – Aaron Lenz Mar 29 at 12:27

$G$ is finite, and you know $H=\{e\}$ satisfies your condition, so you can just take $H$ of maximum order that satisfies this condition. You don't actually have to construct it.


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