1. ## Group Homomorphism?

Let G,G' be groups and φ:G-->G' be a homomorphism. Then prove:
a)if o(g)=n ,then o(φ(g))|n
b)if o(g)=n and φ is injective, then o(φ(g))=n

Firstly I don't actually understand the question. What does the o mean?Is it a new mapping which hasn't been defined or does it represent the composition of functions and if so then how is it computed?Secondly does the |n in (a) represent "divides n" and if so how is that proved, if not then what does it represent? Neither of these appears in my abstract algebra text. In (b) I know that an injective function is one to one but again I don't really understand the question.

Any help would be greatly appreciated. Thanks in advance.

2. ## Re: Group Homomorphism?

o(g) refers to the order of g, which is the smallest positive integer n such that g^n is the identity, or infinity if no such n exists.

3. ## Re: Group Homomorphism?

Originally Posted by chocaholic
Let G,G' be groups and φ:G-->G' be a homomorphism. Then prove:
a)if o(g)=n ,then o(φ(g))|n
b)if o(g)=n and φ is injective, then o(φ(g))=n

Firstly I don't actually understand the question. What does the o mean?Is it a new mapping which hasn't been defined or does it represent the composition of functions and if so then how is it computed?Secondly does the |n in (a) represent "divides n" and if so how is that proved, if not then what does it represent? Neither of these appears in my abstract algebra text. In (b) I know that an injective function is one to one but again I don't really understand the question.

Any help would be greatly appreciated. Thanks in advance.
Yes, $\mid$ means divide. I'll give you the one bit of information that is needed for this problem, namely if $g^k=e$ then $|g|\mid k$ (where I use $|\cdot|$ instead of $o(\cdot)$). From this I think a) should be pretty immediate. For the second, try doing the opposite of the technique you do for the first by proving that $n\mid |\varphi(g)|$.