1. ## Quotient Groups

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

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 $\displaystyle G$ of order $\displaystyle n$ and any $\displaystyle g \in G$, $\displaystyle g^n = e$.

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

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

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

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

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

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

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

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