Question: Let $P$ be an abelian $p$-group with exponent $p^m$. Suppose $M\leq P$ such that $|M|=p^m$, and both $M$ and $P/M$ are cyclic, then there exists $N\leq P$ such that $P=M+N$ and $M\cap N=\{0\}$.
Thoughts: I've wrestled with trying to get the "trivial" cases out of the way first, but, for instance, if $M=P$, then $N$ couldn't even exist, so $M\neq P$. I've tried thinking about the orders of the groups $M$ and $N$ and tried playing with them, but I must not be doing something right, because I am never using that $M$ and $P/M$ are cyclic, which I'm sure (I suppose) are essential in the question... any help is greatly appreciated!
Let $x$ be a generator of $M$, and choose $y \in G$ such that $y+M$ generates $G/M$. Suppose that $|G/M|= p^n$.
Now $p^ny \in M$, and we can write $p^ny = bp^kx$ for some $b$ and $k$ with $\gcd(b,p)=1$.
Then the order of $y$ is the product of $p^n$ and the order of $bp^kx$, which is $p^{m-k}$, so ${\rm ord}(y) = p^{n+m-k}$. But we are told that $G$ has exponent $p^m$, so $n+m-k \le m$ and hence $n \le k$.
So $p^n(y-bp^{k-n}x) = 0$, and hence the subgroup $N$ of $G$ generated by $y-bp^{k-n}x$ satisfies $M+N=G$ and $M \cap N = \{0\}$.