# Abelian $p$-group and proof of the existence of cyclic subgroup

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 $$H=\langle b\rangle\langle c\rangle\cdots$$?

• 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. – M. Vinay Mar 29 at 12:18
• 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? – M. Vinay Mar 29 at 12:19
• @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. – Aaron Lenz Mar 29 at 12:26
• 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. – M. Vinay Mar 29 at 12:27
• @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. – 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.