subgroup

apple2009 · #1 $Old$ November 7th, 2009, 10:43 PM

Let H be a group. Let M be a normal subgroup of H. Let K be any subgroup of H. Let MK={m°k: m is an element of M and k is an element if K}
a)Prove: MK is a subgroup of H.
b)Suppose that M intersects K={I}. Let k, k' are elements of K. Prove M°k=M° k' if and only if k= k'. conclude that |MK|=|M||K|

aliceinwonderland · #2 $Old$ November 7th, 2009, 11:37 PM

Quote:

Originally Posted by apple2009 $View Post$

Let H be a group. Let M be a normal subgroup of H. Let K be any subgroup of H. Let MK={m°k: m is an element of M and k is an element if K}
a)Prove: MK is a subgroup of H.
b)Suppose that M intersects K={I}. Let k, k' are elements of K. Prove M°k=M° k' if and only if k= k'. conclude that |MK|=|M||K|

For (a), it would be helpful to use the following lemma.

Lemma. If M and K are subgroups of a group H, MK is a subgroup of H if and only if MK = KM.

Now you need to show that $MK \subseteq KM$ and $KM \subseteq MK$ in order to apply the above lemma. I'll show $MK \subseteq KM$ and I'll leave it to you to show that the reverse inclusion.

Let $m \in M, k \in K$ such that $mk \in MK$ . Since M is a normal subgroup of H, $mk = k(k^{-1}mk) \in KM$ . Thus $MK \subseteq KM$ .

(b) You might need to use this.
$|MK| = \frac{(|M|)(|K|)}{|M \cap K|}$