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EXERCISE:

Consider that $Φ:G\rightarrow G'$ homomorphism and $a\in G$ which has order d.

Show that the element $φ(a)\in G'$ has also order d.

I am sorry i could't appose any attempt here but i don't even know how to start.I have just started to examine homomorphism and i have to clear my mind on them!

It would be very helpful if someone can give me some hints or a thourough solution to my problem.

I want to learn how i can work with these problems as i don't have many experience with these type of problems!

Thanks in advance!

Community
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Paris K. Patsogiannis
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  • Really interesting exercise ! I think I've seen it while studying for my abstract algebra exams too ! – Rebellos Jan 21 '18 at 20:49
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    This is not true. What is true is that $\operatorname{order}(\Phi(a))\mid d$. –  Jan 21 '18 at 20:52
  • @user8734617 i don't but the exercise is exactly what i have written!So maybe this is what it means! – Paris K. Patsogiannis Jan 21 '18 at 20:54
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    You know $a^d=e_G$ so $\Phi(a)^d=\Phi(a^d)=\Phi(e_G)=e_{G'}$. However, the order of an element $\Phi(a)$ is the smallest number $d'$ such that $\Phi(a)^{d'}=e_{G'}$. $d$ is one such number - but is it the smallest? It doesn't need to be. What is well-known, though, is that $d'\mid d$. For a counterexample, take the trivial homomorphism $\Phi(a)=e_{G'}, a\in G$. Order of $\Phi(a)$ is $1$ for any $a\in G$. –  Jan 21 '18 at 21:01

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To expound on user8734617's comment:

We have that $a^d =e$ (where $e$ is the identity). Therefore $\phi(a)^d = \phi(a^d) = \phi(e) =e$.

However, all this tells us is that the order of $\phi(a)$ divides $d$.

BallBoy
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