Prove that $G$ is cyclic and has exactly two nontrivial subgroups. 1- Let $G$ be an abelian group of order $2p$, where $p$ is an odd prime. Prove that $G$ is
cyclic and has exactly two nontrivial subgroups. Write out in full each theorem or
other previously proven result that you use.?
my try is
I assume $G$ has no nontrivial subgroups.
Because $G$ is trivial by the assumption above, then $G$ is cyclic.
If $G$ has exactly one nontrivial subgroups $H$, consider the subgroup generated by a nonidentity element $g$ in $G/H$
what should I do  next
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2- how you would go about to …find all homomorphic images of a group.?
I know how to find for specific group like $\Bbb Z_4$ or $\Bbb Z_8$ but here not specific any group
how to start
please help me in the two questions
 A: By Cauchy, $G$ has an element of order $2$, say $x$, and an element of order $p$, say $y$. Since $G$ is Abelian, the order of $z:=xy$ is $o(z)=\operatorname{lcm}(2,p)^\dagger=\frac{2p}{\gcd(2,p)}$ ; but $\gcd(2,p)=1$, because $p$ is an odd prime; so,  $o(z)=2p$ and hence $G$ is cyclic.
As for second part of the first point, let $H$ be a nontrivial subgroup of $G$ distinct from $\langle x\rangle$ and $\langle y \rangle$. By Lagrange, $|H|=p$ or $|H|=2$; in the former case, $\langle 
y \rangle H$ has size $p^2$, which is bigger than $2p$ for $p$ odd prime: contradiction; in the latter case, $\langle x\rangle H$ is a subgroup of $G$ of order $4$: contradiction, because $4\nmid 2p$ for $p$ odd prime. Therefore, such a $H$ doesn't exist, and $\langle  x\rangle $ and $\langle  y\rangle $ are the only nontrivial subgroups of $G$.

$^\dagger$By Lagrange, $\langle x\rangle$ and $\langle y\rangle$ intersect trivially, and hence $(xy)^n\stackrel{G\textit{ Abelian}}{=}x^ny^n=e\iff x^n=e\vee y^n=e$.
A: An abelian group whose order is the product of distinct primes is cyclic,  because it is the product of its Sylow subgroups,  which are cyclic and of relatively prime order.
And, a cyclic group has a unique subgroup of each order dividing the order of the group.
