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New to forum, and I was hoping you guys would be able to help me out with this problem. The course is abstract algebra.

The problem statement is:

If f : G → H is a homomorphism, prove the following: Let the range of f have m elements. If a ∈ G has order n, where m and n are relatively prime, then a is in the kernel of f.

I'm at a loss where to begin for this problem.

I know that the range of f is a subgroup of H. I know that the order of an element is defined to be the least positive integer n such that $a^n=e$. I know that the kernel of f is the set K of all the elements of G are carried by f onto the neutral element of H. I just do not know how all these go together to prove the statement.

Thank you!

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  • $\begingroup$ Consider the cyclic subgroup $\langle f(a)\rangle < f(G)$. What can its order be, in light of Lagrange's theorem? $\endgroup$
    – user228113
    Commented Mar 1, 2016 at 7:52
  • $\begingroup$ The order must be Prime? $\endgroup$
    – Miguel
    Commented Mar 1, 2016 at 8:16
  • $\begingroup$ No not necessarily prime. $\endgroup$ Commented Mar 1, 2016 at 8:17
  • $\begingroup$ Fortunately, it is not prime in this case, because the problem asks you to prove otherwise. $\endgroup$
    – user228113
    Commented Mar 1, 2016 at 8:18

1 Answer 1

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Since $a^n=e$, we have $e=f(e)=f(a^n)= f(a)^n$. But, $f(a) \in f[G]$, hence the order of $f(a)$ must divide both $m$ and $n$. Hence $f(a)$ has order one, i.e. $f(a)=e$.

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  • $\begingroup$ I'm sorry for the silly question but can you let me know what f(a)∈f[G] means? f(a)∈H? $\endgroup$
    – Miguel
    Commented Mar 1, 2016 at 8:18
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    $\begingroup$ It means $f(a)\in \{f(x): x\in G\}$. $\endgroup$ Commented Mar 1, 2016 at 8:19

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