3
$\begingroup$

Let $f$ be a homomorphism defined on a finite group $G$, and let $H$ is the subgroup of $G$. Then show that $$ \left [f(G) : f(H)\right] \text{ divides } \left [G : H\right].$$

I know $$\left [G : H\right] = o(G)/o(H);$$

if $o(H) = n$ then $o(G) = kn$ so $o(G)/o(H) = k$.

Likewise $$ \left [f(G) : f(H)\right] = o(f(G)) / o(f(H)).$$

I am stuck here.

Is this the right way of doing this problem?

  • 0
    What is $g$? An element of $G$ would be a fishy thing to say.2012-04-06
  • 2
    Is $f(g)$ a typo for $f(G)$?2012-04-06
  • 0
    @Brian Wouldn't that render this problem trivial? Not that I know of a way to make it non-trivial, : )2012-04-06
  • 0
    @BrianM.Scott NO, f(g) is not a typo for f(G)2012-04-06
  • 0
    @faisal Then what is $g$?2012-04-06
  • 0
    @KannappanSampath g is the element of G. I know its fishy but this is all I have got.2012-04-06
  • 0
    @faisal I have no idea what could be done. Do you know how to solve this exercise if $g$ was $G$ instead? If so, you're doing ok.2012-04-06
  • 0
    If $g\in G$, $[f(g):f(H)]$ really doesn’t make sense as written. Could it be $[\langle f(g)\rangle:f(H)]$, where $\langle f(g)\rangle$ is the group generated by $f(g)$?2012-04-06
  • 0
    @faisal: I believe that there is a typo, otherwise when you get for example $H$ nontrivial, $f$ isomorphism and $g=1$... what happens??2012-04-06
  • 0
    @KannappanSampath No. Tell me if g were G then how would you solve this problem.2012-04-06
  • 0
    Right, at all cost, Brian should have it right. Consider 0o3's comment for a counter example.2012-04-06
  • 0
    @BrianM.Scott you were right. It is f(G).2012-04-06

2 Answers 2