= Solution
<Lagrange's theorem> states that the <group order> of a finite <group> divides the <group order>; more precisely $|G|=[G:H]|H|$.
If every element squares to the <identity element>, every element is its own <inverse element>, and $ab=(ab)^{-1}=b^{-1}a^{-1}=ba$. Thus a <group of exponent two is abelian>. For distinct nonidentity $a,b$, the four elements $1,a,b,ab$ are distinct: for example $ab=1$ would imply $a=b$, and $ab=a$ would imply $b=1$. Products remain in this set because $a,b$ commute and square to one. Thus they form a <Klein four-group> <subgroup> of order four.
Now let $|G|=2p$ with odd <prime number> $p$. By <Lagrange's theorem>, every <element order> is one of $1,2,p,2p$. If an element has order $2p$, its square has order $p$. If no element had order $p$, all nonidentity elements would therefore have order two. There are at least two distinct such elements, producing a subgroup of order four. This contradicts $4\nmid2p$. Hence \b[there is an element of order $p$].
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