1. ## Quotient Groups

Let H be a normal subgroup of finite group G. If the order of the quotient groups $G/H$ is m, prove that $g^m$ is in H for all $g \in G$.

So since H is normal, there are no distinctions between left and right cosets in G.

G is a finite group. Let o(G) = n. The o(G/H) = m.

Don't know where to go after this...

2. Hint: Given any group $G$ of order $n$ and any $g \in G$, $g^n = e$.

In the case of $G/H$, what does this tell you about the coset $g^mH$?

3. Originally Posted by spoon737
Hint: Given any group $G$ of order $n$ and any $g \in G$, $g^n = e$.

In the case of $G/H$, what does this tell you about the coset $g^mH$?
Since $G/H$ is of finite order m, does that imply that $g^mH = eH = H$?

But if $g^mH = H$ then $g^m$ must $\in H$ by theorem (If H is any subgroup of a group G, then hH = H = Hh, $h \in H$).

4. Originally Posted by MissMousey
Since $G/H$ is of finite order m, does that imply that $g^mH = eH = H$?

But if $g^mH = H$ then $g^m$ must $\in H$ by theorem (If H is any subgroup of a group G, then hH = H = Hh, $h \in H$).
Yes, that is correct. $\left|G/H\right|=[G:H]$ and so $\left(gH\right)^{[G:H]}=g^{[G:H]}H=\underbrace{H}_{\text{id elem.}}$ and what you said finishes it.

5. Originally Posted by Drexel28
Yes, that is correct. $\left|G/H\right|=[G:H]$ and so $\left(gH\right)^{[G:H]}=g^{[G:H]}H=\underbrace{H}_{\text{id elem.}}$ and what you said finishes it.
*dances* Thank you!