One way to do this is to look at the group's invariant factor decomposition.
Consider the invariant factor decomposition $\Bbb{F}_{p^n}^{\times} \cong \Bbb{Z}/k_1\Bbb{Z} \oplus \Bbb{Z}/k_2\Bbb{Z} \oplus ... \oplus \Bbb{Z}/k_r\Bbb{Z}$, where $k_{i+1} | k_i$ for all $i$. We wish to show that $\Bbb{F}_{p^n}^{\times}$ is cyclic, or that $r=1$.
The polynomial $x^{k_1} = 1$ must have at most $k_1$ roots. Since, for all $i$, $k_i | k_1$, $x^{k_1} = 1$ for every element $x \in \Bbb{F}_{p^n}^{\times}$. That should be enough for you to find a contradiction if $r>1$.
One possible source of confusion here is that I am writing all groups multiplicatively, using $1$ for the identity, juxtaposition for the group operation and exponents, while the groups $\Bbb{Z}/k\Bbb{Z}$ are usually written additively.
You're fourth step is a bit of an overkill, in terms of testing whether a subset is a subgroup. Indeed, we need to check:
- $(1)$ the identity element of $G$, namely $e = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ is in $H$ and that
- $(2)$ for two elements $h_1, h_2 \in H$, $h_1\cdot h_2 \in H$,
But with respect to testing for inclusion of inverses in H (whether for every element $h \in H$, we also have that $h^{-1} \in H)$, we don't need to test for inverses of products of elements, since you've shown $H$ is closed under multiplication.
Indeed, you are correct that for most $h$ with integer entries $a, b$, the inverse $h^{-1} \notin H$ because the corresponding non-zero entries will likely be rational numbers, but not both integers.
So, taking for example $h = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$, when we compute $h^{-1} = \begin{bmatrix} \frac 12 & 0 \\ 0 & \frac 12\end{bmatrix}$, we've shown that $h^{-1} \notin H$, because $\frac 12 \notin \mathbb Z$.
Hence, we've shown that $H$ is not closed under taking inverses. (All you need to provide is one counterexample of some element $h \in H$ such that $h^{-1} \notin H$, to show that $H$ is not closed under taking inverses.) And we therefore conclude that $H$, as defined, is NOT a subgroup of $G$. And then you're done.
Best Answer
Your proof looks fine to me.
An alternative is this. If $\Bbb Q^\times$ were cyclic, it would be infinite cyclic, so $\simeq \Bbb Z$. But $-1$ has order two in $\Bbb Q^\times$; and there is no element of order two in $\Bbb Z$: every element has infinite order, except for $0$.