If there exists a homomorphism from $G$ to $G'$, if there exists an element $x$ of order $n$ in $G'$, then we must have an element of order $n$ in $G$
What if the homomorphism is onto?
Counter example of the first statement : Take the trivial homomorphism from $Z_2 $ to $Z_8$
I'll try to prove the onto case :
Let $x\in G', \; |x|=n \Rightarrow \exists y\in G \; \text{s.t} \; \phi(y)=x \Rightarrow |x| \; \text{divides} \; |y| \Rightarrow |y|=nk \; \text{f.s} \; k \in N \Rightarrow |y^{k}| = n $ and so $y^k$ is required element of $G$
Is this correct?