# iso and cyclic

• Aug 3rd 2008, 06:22 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, 06:58 AM
PaulRS
Quote:

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

$G_1$ 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
$
• Aug 3rd 2008, 09:25 AM
JaneBennet
Quote:

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 $n\in\mathbb{Z}$

$\Rightarrow\ x=f(a^n)=\left[f(a)\right]^n\in\left$
• Aug 3rd 2008, 09:58 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, 06:36 PM
algebraic topology
Quote:

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.
• Aug 5th 2008, 09:29 AM
JaneBennet
Quote:

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. (Clapping)