- #1

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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

- Thread starter SquareCircle
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- #1

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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

matt grime

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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)).

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)).

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- #3

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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:

- #4

matt grime

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

- #5

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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

matt grime

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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

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Thank you, your explanation of the proof helped.

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