# Show that if $G$ is cyclic then so is $H$

If group $G$ is isomorphic to group $H$, show that if $G$ is cyclic then so is $H$.

An isomorphism is simply a bijective homomorphism. The latter is a function which preserves the group operation. $f(g_1*g_2)= f(g_1) \cdot f(g_2)$, $g_1, g_2 \in G$

A cyclic group is a group generated by one element. I'm not sure how to connect these though

• Suppose $G$ is generated by $g$, what can you say about $f(g)$? – Cameron Williams Oct 30 '14 at 1:29
• Hint: $f(g)^n=f(g^n)$, use this to show that $f(g)$ has the same order as $g$. – Peter Huxford Oct 30 '14 at 1:31
• @Peter your initial part makes sense since $f$ is a isomorphism. But how do I show $f(g)$ has the same order as $g$? – atherton Oct 30 '14 at 1:34
• @atherton let $k$ be the order of $f(g)$, i.e. the smallest positive integer such that $f(g)^k = e_H$, where $e_H$ is the identity in $H$. Then $f(g^k)=e_H=f(e_G)$, which means that $g^k=e_G$. This shows that $k\geq n$, where $n$ is the order of $g$. Additionally $f(g)^n=f(g^n)=f(e_G)=e_H$, so $n\geq k$. Therefore the order $n$ and $k$ are equal. – Peter Huxford Oct 30 '14 at 1:47

If $G=\langle g\rangle$ then $H=\langle f(g)\rangle$.
Let $h\in H$ then there exists $x=g^k\in G$ such that $h=f(x)=f(g^k)=f(g)^k$.