# Thread: iso and cyclic

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
$\displaystyle G_1$ is cyclic iff there exists $\displaystyle a\in G_1$ such that $\displaystyle \left\langle a \right\rangle = G_1$. Okay, we now say, first, since there's a homomorphism: $\displaystyle f\left( {a^n } \right) = f\left( a \right)f\left( {a^{n - 1} } \right) = ... = f^{n } \left( a \right)$ (1)

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

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

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

If $\displaystyle G_1=\left<a\right>$ is cyclic and $\displaystyle f:G_1\to G_2$ is an isomorphism, you just have to show that $\displaystyle G_2=\left<f(a)\right>$. Similarly, if $\displaystyle G_2=\left<b\right>$ is cyclic, then $\displaystyle G_1=\left<f^{-1}(b)\right>$.

Thus, for the first part:

We have that $\displaystyle \left<f(a)\right>\subseteq G_2$ so we only need to show that $\displaystyle G_2\subseteq \left<f(a)\right>$.

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

$\displaystyle \Rightarrow\ f^{-1}(x)\in G_1=\left<a\right>$

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

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

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 $\displaystyle \left<f(a)\right>\subseteq G_2$ so we only need to show that $\displaystyle G_2\subseteq \left<f(a)\right>$.

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

$\displaystyle \Rightarrow\ f^{-1}(x)\in G_1=\left<a\right>$

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

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

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