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

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- Aug 3rd 2008, 05:22 AMmathemanyakiso and cyclic
let f: G1->G2 be an isomorphism then show that G1 is cyclic iff G2 is cyclic

- Aug 3rd 2008, 05:58 AMPaulRS
$\displaystyle G_1$ 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

$ - Aug 3rd 2008, 08:25 AMJaneBennet
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 $\displaystyle n\in\mathbb{Z}$

$\displaystyle \Rightarrow\ x=f(a^n)=\left[f(a)\right]^n\in\left<f(a)\right>$ - Aug 3rd 2008, 08:58 AMNonCommAlg
**JaneBennet**is right but, considering the fact that**PaulRS**is only 18 years old, it's fascinating that how knowledgeable he is! (Clapping) - Aug 3rd 2008, 05:36 PMalgebraic topology
- Aug 5th 2008, 08:29 AMJaneBennet