# iso and cyclic

• Aug 3rd 2008, 05:22 AM
mathemanyak
iso and cyclic
let f: G1->G2 be an isomorphism then show that G1 is cyclic iff G2 is cyclic
• Aug 3rd 2008, 05:58 AM
PaulRS
Quote:

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 $\displaystyle f^{ - 1} :G_2 \to G_1$
• Aug 3rd 2008, 08:25 AM
JaneBennet
Quote:

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 $\displaystyle n\in\mathbb{Z}$

$\displaystyle \Rightarrow\ x=f(a^n)=\left[f(a)\right]^n\in\left<f(a)\right>$
• Aug 3rd 2008, 08:58 AM
NonCommAlg
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 PM
algebraic topology
Quote:

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.
• Aug 5th 2008, 08:29 AM
JaneBennet
Quote:

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. (Clapping)