Proposition 1: $(\mathbb{Z}/p^{a}\mathbb{Z})^{\times}$ is cyclic group of order $p^{a-1}(p-1)$, where $(\mathbb{Z}/p^{a}\mathbb{Z})^{\times}$ is the multiplicative group of $\mathbb{Z}/p^{a}\mathbb{Z}$ of unit elements
It's easy to see that $|(\mathbb{Z}/p^{a}\mathbb{Z})^{\times}| = \varphi(p^{a})=p^{a-1}(p-1)$.
Proposition 2: A finite subgroup of the multiplicative group of a field is cyclic.
Using proposition 2, I can prove proposition 1 as follows: Let $F$ be the field of fraction of $\mathbb{Z}/p^{a}\mathbb{Z}$. Then $(\mathbb{Z}/p^{a}\mathbb{Z})^{\times}$ is a finite subgroup of of multiplicative group of field $F$, by proposition 2 we have $(\mathbb{Z}/p^{a}\mathbb{Z})^{\times}$ is cyclic.
I think the proof above should works for all primes (not only just odd primes). However, when $p=2$, the group $(\mathbb{Z}/2^{a}\mathbb{Z})^{\times}$ is not cyclic, in fact, it's the direct product of a cyclic group of order 2 and a cyclic group of order $2^{a-2}$.
Where did my proof go wrong in the case $p=2$? Any helps would be appreciated! Thanks.