1. ## iso and cyclic

let f: G1->G2 be an isomorphism then show that G1 is cyclic iff G2 is cyclic

2. Originally Posted by mathemanyak
let f: G1->G2 be an isomorphism then show that G1 is cyclic iff G2 is cyclic
$G_1$ is cyclic iff there exists $a\in G_1$ such that $
\left\langle a \right\rangle = G_1
$
. Okay, we now say, first, since there's a homomorphism: $
f\left( {a^n } \right) = f\left( a \right)f\left( {a^{n - 1} } \right) = ... = f^{n } \left( a \right)
$
(1)

Since $
\left\langle a \right\rangle = G_1
$
it must be because $
G_1 = \left\{ {a,a^2 ,...,a^{\left| G_1 \right|} } \right\}
$
(the order of a is |G_1|) being all those elements different. $f$ is biyective thus: $G_2=
f\left( G_1 \right) = \left\{ {f\left( a \right),f\left( {a^2 } \right),...,f\left( {a^{\left| G_1 \right|} } \right)} \right\}
$
by (1): $
G_2 = \left\{ {f\left( a \right),f\left( {a^2 } \right),...,f\left( {a^{\left| {G_1 } \right|} } \right)} \right\} = \left\{ {f\left( a \right),f^2 \left( a \right),...,f^{^{\left| {G_1 } \right|} } \left( a \right)} \right\}
$
thus $
G_2 = \left\langle {f\left( a \right)} \right\rangle
$

The converse is analogous, take $
f^{ - 1} :G_2 \to G_1
$

3. Originally Posted by PaulRS
$G_1 = \left\{ {a,a^2 ,...,a^{\left| G_1 \right|} } \right\}$
You are assuming that $G_1$ is finite. The question doesn’t say anything about finite groups, though.

If $G_1=\left$ is cyclic and $f:G_1\to G_2$ is an isomorphism, you just have to show that $G_2=\left$. Similarly, if $G_2=\left$ is cyclic, then $G_1=\left$.

Thus, for the first part:

We have that $\left\subseteq G_2$ so we only need to show that $G_2\subseteq \left$.

$\color{white}.\quad.$ $x\in G_2$

$\Rightarrow\ f^{-1}(x)\in G_1=\left$

$\Rightarrow\ f^{-1}(x)=a^n$ for some $n\in\mathbb{Z}$

$\Rightarrow\ x=f(a^n)=\left[f(a)\right]^n\in\left$

4. JaneBennet is right but, considering the fact that PaulRS is only 18 years old, it's fascinating that how knowledgeable he is!

5. Originally Posted by JaneBennet
Thus, for the first part:

We have that $\left\subseteq G_2$ so we only need to show that $G_2\subseteq \left$.

$\color{white}.\quad.$ $x\in G_2$

$\Rightarrow\ f^{-1}(x)\in G_1=\left$

$\Rightarrow\ f^{-1}(x)=a^n$ for some $n\in\mathbb{Z}$

$\Rightarrow\ x=f(a^n)=\left[f(a)\right]^n\in\left$
This shows that $f:G_1\to G_2$ only needs to be a surjective homomorphism in order for $G_1$ cyclic $\implies$ $G_2$ cyclic to hold.

6. Originally Posted by algebraic topology
This shows that $f:G_1\to G_2$ only needs to be a surjective homomorphism in order for $G_1$ cyclic $\implies$ $G_2$ cyclic to hold.
Hence, if $f$ is an isomorphism, the fact that $f^{-1}$ is a surjective homomorphism from $G_2$ onto $G_1$ proves the reverse implication and completes the proof.