Let $N$ be a normal subgroup of finite group $G$. Prove that if order of $H$ and order of $G/N$ are relatively prime then $H$ is a subgroup of $N$.

Question

Let $N$ be a normal subgroup of finite group $G$. Prove that if order of $H$ and order of $G/N$ are relatively prime then $H$ is a subgroup of $N$.

Can someone help me to understand the intuition behind this question and why is this happening with example?

I am basically clueless about how to start.

What is the order of $\pi(H)$, where $\pi:G\to G/N$ is the projection? — user722227, Jan 27 '20 at 14:58

score 2 · Accepted Answer · 2020-01-27T15:39:14.477

2

By the first isomorphism theorem, the order of a homomorphic image of a group divides the order of the group.

Consider the canonical projection $π:G\to G/N$. Then $|π(H)|\mid|H|$.

And by Lagrange, $|π(H)|\mid|G/N|$.

So $π(H)=0$. So $H\le N$.

edited Jan 27 '20 at 15:39

answered Jan 27 '20 at 15:27

We have not been taught the theorems of isomorphism yet so I presume that we are supposed to solve it without using isomorphism theorems.. – Antimony Jan 27 '20 at 16:30
Have a look at https://math.stackexchange.com/a/1937048. – Jan 27 '20 at 19:49

1 Answers1