Let G be a group, and H a normal subgroup of G. Prove that, for each element a in G, the order of the element Ha in G/H is a divisor of the order of a in G.
So I have already done a lot of stuff with homomorphisms and the order of elements, and I know G/H is a homomorphic image of G. I'm just not sure how I can apply this to quotient groups.
an = e for some n in G. Does this mean that I'm looking for some Hn = e? I guess I'm having a hard time thinking of Ha as an element.
Best Answer
For quotient group $G/H$, you need to remember that its element is $Hg$, where $g\in G$, and $He=H$ is the identity of $G/H$. The multiplication is $(Hg)(Hg_1)=Hgg_1$. Now let $m$ be the order of $a$, then $(Ha)^m=Ha^m=H$, this means the order of $Ha$ is a divisor of $m$.