= Solution
<Lagrange's theorem> states that, for a <finite group> $G$ and a <subgroup> $H$,
$$
\boxed{|G|=[G:H]|H|.}
$$
To prove it, define an equivalence relation by $x\sim y$ when $x^{-1}y\in H$. Its classes are the left <cosets> $xH$, which therefore partition $G$. Left multiplication by $x$ gives a <bijection> $H\to xH$, so every <coset> has $|H|$ elements. Counting this partition gives the formula; in particular $|H|$ divides $|G|$. Applying the theorem to the <cyclic subgroup> generated by an element shows that its <order of a group element> divides $|G|$.
Write the given order-two <normal subgroup> as $N=\{1,z\}$. Conjugation must preserve $N$ and cannot send its nonidentity element to the identity, so $gzg^{-1}=z$ for all $g\in G$. Thus $z$ is in the <center of a group>: a <normal subgroup of order two is central>.
Choose $g\notin N$. By <Lagrange's theorem>, its <order of a group element> is $2$, $p$ or $2p$. It cannot have order two: since $z$ is central, the four distinct elements $1,z,g,zg$ would form a <subgroup> of order four, whereas $4$ does not divide $2p$ for odd $p$. If $g$ has order $2p$, it already generates $G$. Otherwise $g$ has order $p$, commutes with $z$, and $(gz)^k=1$ implies $g^k=z^{-k}\in\langle g\rangle\cap N=\{1\}$. Hence both $p$ and $2$ divide $k$, so $gz$ has order $2p$. \b[In either case]
$$
\boxed{G\cong C_{2p}.}
$$
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