If $[G:K] = m$, then prove that $g^m \in K$ for all $g \in G$.

Question

Let $K$ be a normal subgroup of $G.$ If $[G:K] = m$ then prove that $g^m \in K$ for all $g \in G$.

Solution: If $[G:K] = m$ then $|G/K| = m$. By Lagrange's theorem, the order of an element divides the order of the group. Therefore $(gK)^m = K$ for any coset $gK$. So for any $g \in G$, we have $(gK)^m = g^mK = K$, which means that $g^m \in K$. $\square$

I don't understand why we can make the claim that $(gK)^m = K$ for all cosets $gK$ because the order of an element divides the order of the group.

The order of an element $x$ divides an integer $n$ iff $x^n$ is the identity. — azif00, Apr 16 '23 at 21:51
...and the order of the element $gK\in G/K$ divides the order $[G:K]$ of the group $G/K$ (whose identity element is $K$). — Anne Bauval, Apr 16 '23 at 21:55
Two people cooperating on completing an answer in the comment section today. That's not where answers should go. — Arthur, Apr 16 '23 at 21:57
@AnneBauval so what you guys are saying is that in every single group, every single element to the power of the order of the group will give back the identity — user20194358, Apr 16 '23 at 21:59
@user20194358 If $G$ is a finite group then by Lagrange's theorem the order of every element $g\in G$ divides $|G|$. It follows that $g^{|G|}=e$. — Mark, Apr 16 '23 at 22:04
See also https://math.stackexchange.com/questions/573050/if-h-is-a-subgroup-of-g-of-finite-index-n-then-under-what-condition-gn — lhf, Apr 16 '23 at 23:56

score 3 · Accepted Answer · answered Apr 16 '23 at 23:09

3

Since $K$ is normal in $G$, $G/K=\{gK|\forall g\in G\}$ is the quotient group (factor group), the order of this quotient group is $m$. Now, take an arbitray element $gK$ from this quotient group, we have $(gK)^m=K$. Note $K$ plays as the unit element in the quotient group (like $e$ plays as the unit element in $G$). Because $K$ is normal in $G$, which means $gK=Kg$, so we have $(gK)^m=g^mK^m=g^mK=K$, and this implies $g^m\in K.$

answered Apr 16 '23 at 23:09

MathFail

21,128

1

This does not answer the question. The OP wants to know why $g^mK=K$. Please read the question more carefully. – Shaun Apr 16 '23 at 23:11
3

"Note K plays as the unit element in the quotient group (like e plays as the unit element in G)." Does this answer it? – MathFail Apr 16 '23 at 23:14
Ah, I see; sorry! – Shaun Apr 16 '23 at 23:15
no worries. :-) – MathFail Apr 16 '23 at 23:16

If $[G:K] = m$, then prove that $g^m \in K$ for all $g \in G$.

1 Answers1