Complement of a cyclic subgroup in a finite, abelian p-Group (Proof-Verification) I'm trying to prove following statement, can anyone check if my proof works? In Isaacs, I.M. Algebra - A Graduate Course the proof is really long and complicated(but needs less theory), so I'm afraid I made a mistake, as the statement seems to be quite trivial with the use of the fundamental theorem of finitely generated abelian groups.
Lemma 1
In a finite, abelian p-group $G$, a cyclic subgroup $H$ of maximal possible order has a complement.
The following proof is incomplete, see my answer for a correct one.
Proof. Let $G\cong \Bbb{Z}_{p^{r_1}}\oplus\cdots \oplus \Bbb{Z}_{p^{r_s}}$ with $r_1 = \cdots =r_l \geq  ... \geq r_s$.
As $H$ is cyclic and of maximal order, $H=\Bbb{Z}_{p^{r_i}}$ for $i\in\{1,\cdots,l\}$.
Let $X$ be the subgroup consisting of all summands, but the $i$-th. Meaning $X=\bigoplus \Bbb{Z}_{p^{r_k}}$, $k \in \{1,\cdots,s\} \setminus \{i\}$.
Then $G=H \oplus X$ and $H$ has a complement in $G$.
q.e.d.
 A: As I see it, your proof is incomplete.
The claim to be proved is this:

*
In a finite abelian $p$-group, every cyclic subgroup of maximum order has a complement.

You chose a cyclic subgroup of maximum order from among those of maximum order in a particular direct sum decomposition. But there can be others.

In fact, if a given direct sum decomposition has more than one summand, there will be other cyclic subgroups of maximum order besides those which appear in the direct sum decomposition. 
A: Thanks to quasi I saw that my proof is incomplete. Here is now a full proof:
Lemma 1
In a finite, abelian p-group $G$, a cyclic subgroup $H$ of maximal possible order has a complement.
Proof.
If $G$ is cyclic, then $H=G$ and nothing needs to be shown. Let $G$ be not cyclic.
We make an induction over $|G|$.
If $|G|=p$, then $G$ is cyclic and we are finished.  Let $|G|>p$, and suppose, our lemma is true for all groups with order smaller then $|G|$.
Because $G$ is not cyclic, there is more than one subgroup of order $p$. Let $K \leq G$ be a subgroup with $|K|=p$ and $K \not \leq H$. $K$ has prime order, hence $H \cap K = \{e\}$ and $(HK)/K \cong H$.
For arbitrary $g \in G$ with corresonding coset $gK \in G/K$ is the following true: $ord(gK) | ord(g) \leq |H|$. Though $(HK)/K$ is a subgroup of maximal order in $G/K$. Our induction-hypothesis implies the existence of a complement $\bar{H'} \leq G/K$ of $(HK)/K$ in $G/K$. $\bar{H'}$ is the image of a group $H'$ under the canonical epimorphism $G \rightarrow G/K$ with $K \leq H \leq G$(correspondence-theorem).
Hence $G/K=((HK)/K)(H'/K)=(HKH')/K$ and finally: $$G=(HK)H'=H(KH')=HH' \; \text{with} \; H \cap H' = \{e\}.$$
