# Homomorphism of a cyclic subgroup is a cyclic subgroup ?

1. Mar 20, 2012

### Leb

1. The problem statement, all variables and given/known data

Let $\alpha:G \rightarrow H$ be a homomorphism and let x$\in$G

Prove $\alpha(<x>) =<\alpha(x)>$

2. Relevant equations

α(<x>) = α({x^{r}: r ∈ Z}) = {α(x^{r}) : r ∈ Z} = {α(x)^{r}: r ∈ Z} = <α(x)>.

I do not understand how can we take out the 'r' out of a(x^{r}) to get (a(x))^{r} ? What would a(x)^2 say mean ? a(x) (some binary operation defined in H) a(x) or would it be (a°a)(x) i.e. a(a(x)) ?

2. Mar 20, 2012

### tiny-tim

Hi Leb!

(try using the X2 button just above the Reply box )
No, that would be ar(x).

This is (a(x))r, the product of r elements of group H.

3. Mar 20, 2012

### Leb

However, I am still not sure what will happen...

(a(x))r = a(x) * a(x) *.....*a(x) r-times ? But what is * then ? Is it defined in H or in G ?

4. Mar 20, 2012

### Karamata

Well, you have that $a$ is homomorphism, so $a\bigl(x^2\bigr)=a(x\star x)=a(x)*a(x)=\bigl(a(x)\bigr)^2$, same for $a(x^{r})$, where $\star$ is binary operation in $G$ and $*$ in $H$.

5. Mar 20, 2012

### tiny-tim

in H

x is in G, a(x) is in H …

so x can only undergo G's operations,

and a(x) can only undergo H's operations

6. Mar 20, 2012

### Leb

Ah, OK, now it is clear. Thanks guys !