If $N$ is a normal subgroup of $G$ and has an index of 12 in $G$, then what can I say about the order of elements in $N$ or in $G$?

Question

I'm quite new to the concept of index of a subgroup but I know the relation

$[G:H] = \frac{|G|}{|H|}$ but this relation tells me the order of the group, not it's elements. What am I missing here. Is any of the following options true?

(a) $x^{12}=e$ for all $x∈N$

(b) $x^{12}=e$ for all $x∈G$

(c) $x^{24}∈N$ for all $x∈G$

Answer 1

If $N$ is a normal subgroup of $G$ and has an index of $12$ in $G$, then what can I say about the order of elements in $N$ or in $G$?

Not much. Here's an example $G=N\oplus\mathbb{Z}_{12}$. Clearly $[G:N]=12$ and $N$ is normal in $G$ yet $N$ was chosen arbitrarly meaning elements can have any possible order. Meaning (a) and (b) are not true.

Now for (c): since $N$ is normal then $G/N$ is a group of order $12$. It follows that for any $x\in G$ we have

$$(xN)^{12}=N$$

i.e. $x^{12}N=N$ and thus $x^{12}\in N$ for any $x\in G$. Or more generally $x^{[G:N]}\in N$. So its a variant of (c) except that it is $12$, not $24$.

Answer 2

Hint: apply Lagrange’s Theorem in the quotient group $G/N$. And you will see that (c) holds.