Order of automorphism group of a $p$-group

Question

This is basically a question regarding a solution from this old topic Order of automorphism group of a $p$-group is divisible by $p$. to the abelian case.
In the only answer of this question, two cases are considered. I can understand the first case, but there is something about the second one which I have a hard time understanding. Why can we say that $G \cong \mathbb{Z_{p^k}} \times H$ where $k\ge 2$?. From the classification theorem of finite abelian groups we just know that $G \cong\mathbb{Z_{p^{\alpha_1}}}\times \mathbb{Z_{p^{\alpha_2}}}\times...\times \mathbb{Z_{p^{\alpha_n}}}$, where $1\le \alpha_1\le \alpha_2 \le...\le \alpha_n$ are integers. I understand the fact that at least one of these integers is greater than $1$ in this case, but to me it looks like we are assuming that $\alpha_1 >1$.

Answer 1

It was not said there that $\alpha_1>1$, but rather, that some of the $\alpha_i$ is greater than 1. The argument in the accepted answer works if any of the direct summands has order $p^k$ with some $k>1$.