Proving an Isomorphism

1. Feb 23, 2005

SquareCircle

How would I go about proving the following:

If G has an element of order n, then H has an element of order n.

I am not sure how to start, if I should some how go about proving one to one and onto.

Help

2. Feb 23, 2005

matt grime

Who knows, since you've not explained what G and H are.

But, guessing you mean let G and H be isomorphic groups, show that G has an element of order n iff H does.

Suppose f is an iso from G to H. Let x be in G, then, f(x^r)=f(x)^r, hence ord(f(x))<=ord(x). by symmetry ord(x)=ord(f(x)).

Last edited: Feb 23, 2005
3. Feb 23, 2005

SquareCircle

Isomorphism

Sorry, I left that part out.

The whole problem states

Assume that G and H are groups and that G and H are isomorphic
Then prove the statement
If G has an element of order n, then H has an element of order n.

Last edited: Feb 23, 2005
4. Feb 24, 2005

matt grime

Which is what I showed, albeit in a very quick fashion. Do you understand the proof?

5. Feb 24, 2005

SquareCircle

Isomorphism

No, I do not understand the proof. I am taking group theory and I do not understand the concepts. Do you know what I can do to help me understand some of the concepts?

6. Feb 24, 2005

matt grime

The concept is simply a definition.

the order of an element is the smallest positive r such that x composed with itself r times is the identity

a group isomorphism is a structure preserving map f(xy)=f(x)f(y)

so it follows f(x^r)=f(x)^r

if x^r=e, the identity, then f(x)^r = e, so if r is minimal and positive such that x^r = e then f(x) has order at most r. So by symmetry, with g the inverse iso to f, it follows they must be equal.

you need to think about it. it shouldn't be instantly obvious, it'll take time to understand, but it's supposed to

7. Feb 24, 2005

SquareCircle

Isomorphism

Thank you, your explanation of the proof helped.