A normal subgroup $H$ of $G$, $[G:H] = n$

Question

Let $H\triangleleft G$, and H is a subgroup of G such that $[G:H] = k$, where $[G:H]$ states the number of left cosets of $H$ in $G$.

1) How may I show that for all $a\in G, a^k\in H$?

2) If we do NOT assume the normality of $H$, can I have any counterexample showing that 1) is not true?

One typically says "is a subgroup of index $k$", without the "the". — Daniel Fischer, Nov 03 '16 at 16:25
Concerning question 2), look at the smallest group having a non-normal subgroup. Look at a non-normal subgroup of that. — Daniel Fischer, Nov 03 '16 at 16:25
Related: https://math.stackexchange.com/questions/472672, https://math.stackexchange.com/questions/510023 — Watson, Nov 03 '16 at 16:29

score 1 · Accepted Answer · answered Nov 03 '16 at 16:19

1

I assume, $[G:H]=k$, then the order of $G/H$ is $k$ and $\bar a^k=1$ where $\bar a$ is the class of $a$ in $G/H$ and $a^k\in H$, since its class in $G/H$ is $1$.

answered Nov 03 '16 at 16:19

Tsemo Aristide

87,475

A normal subgroup $H$ of $G$, $[G:H] = n$

1 Answers1