= Solution
If a nonidentity power $x^j$ belongs to $H$, reduce $j$ modulo $p$ so that $1\le j<p$. There is an integer $a$ with $aj\equiv1\pmod p$, and therefore $(x^j)^a=x\in H$, a contradiction. Thus
$$
\boxed{\langle x\rangle\cap H=\{1\}.}
$$
Now suppose that $G$ is abelian and finite. Start with $H_0=\{1\}$, and whenever $H_k\ne G$, choose $x_{k+1}\notin H_k$. The multiplication map
$$
H_k\times\langle x_{k+1}\rangle\longrightarrow H_{k+1}=H_k\langle x_{k+1}\rangle
$$
is a <group homomorphism> because all elements commute. It is surjective by construction and injective because the intersection of its two factors is trivial: $hx=h'x'$ implies $h'^{-1}h=x'x^{-1}$ lies in that intersection. Hence $H_{k+1}\cong H_k\times C_p$, and its order is $p$ times the previous order. Strict growth and finiteness ensure that the construction ends at $G$. Therefore
$$
\boxed{G\cong C_p^n,}
$$
where $n$ is the number of factors, possibly zero. This is an <elementary abelian group>, obtained by an explicit <direct product of groups> construction.
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