Quotient Groups

**MissMousey** · May 2nd 2010, 06:47 PM

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...

**spoon737** · May 2nd 2010, 08:35 PM

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 $g^mH$ ?

**MissMousey** · May 4th 2010, 02:14 PM

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, $h \in H$ ).

**Drexel28** · May 4th 2010, 03:25 PM

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.

**MissMousey** · May 4th 2010, 03:30 PM

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!

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