isomorphic cyclic groups

lllll · #1 $Old$ November 3rd, 2009, 08:03 PM

I'm having a lot of trouble showing that:

$Z_2 \times \mathbb{Z} \ \cong \ \mathbb{Z}$

I figured the simplest way to go about it was to show that there is no bijection between both groups, but am clueless on how to go about it.

I realize that the only elements of $Z_2$ are {1,x} such that $x^2=1$

and $\mathbb{Z}$ is just $\{0,\pm 1, \pm 2,... \pm z, ...\}$ but I'm not sure on how to represent $Z_2 \times \mathbb{Z}$

tonio · #2 $Old$ November 3rd, 2009, 09:04 PM

Quote:

Originally Posted by lllll $View Post$

I'm having a lot of trouble showing that:

$Z_2 \times \mathbb{Z} \ \cong \ \mathbb{Z}$

I figured the simplest way to go about it was to show that there is no bijection between both groups, but am clueless on how to go about it.

I realize that the only elements of $Z_2$ are {1,x} such that $x^2=1$

and $\mathbb{Z}$ is just $\{0,\pm 1, \pm 2,... \pm z, ...\}$ but I'm not sure on how to represent $Z_2 \times \mathbb{Z}$

I suppose you meant that you have to show that these two groups are NOT isomorphic, and the proof is simple: $\mathbb{Z}_2\times \mathbb{Z}$ has a non-trivial element of finite order whereas $\mathbb{Z}$ hasn't.

Tonio

lllll · #3 $Old$ November 3rd, 2009, 10:43 PM

Quote:

Originally Posted by tonio $View Post$

$\mathbb{Z}_2\times \mathbb{Z}$ has a non-trivial element of finite order whereas $\mathbb{Z}$ hasn't.

but how does the fact that one group having an element of non-trival order mean that there is no isomorphism between the a group that doesn't?

Does it mean that there is z in $\mathbb{Z}_2\times \mathbb{Z}$ such that z generates the entire group?

tonio · #4 $Old$ November 4th, 2009, 04:21 AM

Quote:

Originally Posted by lllll $View Post$

but how does the fact that one group having an element of non-trival order mean that there is no isomorphism between the a group that doesn't?

Does it mean that there is z in $\mathbb{Z}_2\times \mathbb{Z}$ such that z generates the entire group?

Not at all: $\mathbb{Z}_2\times \mathbb{Z}$ is not cyclic, because it if were then it would either be a cyclic of finite order, which it isn't since it contains $\mathbb{Z}$ , or else it'd be infinite cyclic and thus isomorphic to $\mathbb{Z}$ which, as said, it is not.

Remember that isomorphisms preserve orders of elements, and this follows from the next fact that I invite you to prove (it's easy):

Lemma: if $\phi :G\rightarrow H$ is a group homomorphism, then $\forall\,g\in G\,,\,\,ord(\phi(g))\mid ord(g)$ , and if g is of infinite order then $\phi(g)$ is either of infinite order or the trivial element in H.

Thus, to show that an isomorphism preserve orders apply the above lemma twice: to the isomorphism and to its inverse isomorphism.

Tonio