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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

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    $\begingroup$ Suppose $G$ is generated by $g$, what can you say about $f(g)$? $\endgroup$ – Cameron Williams Oct 30 '14 at 1:29
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    $\begingroup$ Hint: $f(g)^n=f(g^n)$, use this to show that $f(g)$ has the same order as $g$. $\endgroup$ – Peter Huxford Oct 30 '14 at 1:31
  • $\begingroup$ @Peter your initial part makes sense since $f$ is a isomorphism. But how do I show $f(g)$ has the same order as $g$? $\endgroup$ – atherton Oct 30 '14 at 1:34
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    $\begingroup$ @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. $\endgroup$ – Peter Huxford Oct 30 '14 at 1:47
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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 $h=f(x)=f(g^k)=f(g)^k$.

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