0
$\begingroup$
  1. Let G be a finite group with a normal subgroup N. Prove that if K is a subgroup of $G$ such that $K ∩ N = {e}$ then $|K||[G : N]$
  2. Let G be a cyclic group and $f : G → G$ a function. Prove that $f$ is a homomorphism iff there is an integer $k$ such that $f(a) = a^{k}$ for every $a ∈ G$.

For the first question would the second iso thm be used here? Since $KN/N$ isomorphic to $K/K ∩ N$ can be reduced to by lagrange's thm |KN/N|=|K| but I am not sure what I would do after this.

For the second question this looks familar to the defintion of a subgroup of a cyclic group but I am not sure how to approach it..

$\endgroup$

3 Answers 3

1
$\begingroup$

Since $N$ is normal in $G$ the set $KN$ is a subgroup of $G$. Now, $|KN|=\frac{|K||N|}{|K\cap N|}=|K||N|$. Now the index of the subgroup $KN$ in $G$ is $$[G:KN]=\frac{|G|}{|KN|}=\frac{1}{|K|}\frac{|G|}{|N|}\implies |K|[G:KN]=[G:N]\implies|K|\bigg| [G:N].$$

Next if $f$ is a group homomophism then $f(a)=a^k$ for each $a\in G$ as $G=\langle g\rangle\implies f(g)=g^n$ for some $n\in \Bbb Z$. So that, $f(g^k)=g^{nk}$ for all $k\in\Bbb Z$.

Now suppose we have a fixed integer $k$ for which, $f(a)=a^{k}$ for each $a\in G$. Then clearly $f(a\cdot b)=(ab)^k=a^kb^k=f(a)f(b),\forall a,b\in G$ as cyclic groups are abelian. So $f$ is a group homomorphism.

$\endgroup$
1
$\begingroup$

By the third isomorphism theorem, you have that every subgroup of $G/N$ is of the form $H/N$, for some subgroup $H$ in $G$.

So all you need to prove is that $KN$ is a group itself (from which it follows it is a subgroup of $G$, not just a subset). You might know already that $KN$ is a group of $G$ iff $KN=NK$, and this follows since $N$ is a normal subgroup.

You therefore have the required result by Lagrange's theorem.

$\endgroup$
0
$\begingroup$

Hint: For the second question, write $f(g)=g^k$, where $G=\langle g \rangle$.

$\endgroup$

You must log in to answer this question.