1. ## normal subgroup question

Suppose that H is a normal subgroup of G. If | H|=4 and gH has order 3 in G/H, find a subgroup of order 12 in G.
How would know what the subgroup of order 12 in G would look like?

2. Originally Posted by nhk
Suppose that H is a normal subgroup of G. If | H|=4 and gH has order 3 in G/H, find a subgroup of order 12 in G.
How would know what the subgroup of order 12 in G would look like?

Look at the subgroup $\left\leq G\slash H\Longrightarrow \left=K\slash K$ , by the correspondence theorem for some subgroup $H\leq K\leq G$ .

Assuming the group $G$ is finite, we get $3=|\left|=\left|K\slash H\right|=\frac{|K|}{|H|}\longrightarrow |K|=3|H|=12$

Tonio

3. Can you explain to me where you get the K subgroup and how it plays into your proof? That would be very helpful.

4. Let K=. Since |H|=4 and gH has order 3 in G/H, |K|=12. Lets now prove that K is a subgroup by the two step subgroup test.
Obviously K is a subset of H.
How would my elements look when I am doing a two step subgroup test when proving that the identity is in K, that the element ab is in K whenever a and b are in K, and a-1 is in K whenever a is in K

5. Originally Posted by nhk
Let K=. Since |H|=4 and gH has order 3 in G/H, |K|=12. Lets now prove that K is a subgroup by the two step subgroup test.
Obviously K is a subset of H.

No, rather obviously it is not: for example, $g\in gH\subset K$ , but $g\notin H$ ...I'm assuming you meant to write $gH$ instead of $GH$.

Tonio

How would my elements look when I am doing a two step subgroup test when proving that the identity is in K, that the element ab is in K whenever a and b are in K, and a-1 is in K whenever a is in K
.

6. Originally Posted by nhk
Can you explain to me where you get the K subgroup and how it plays into your proof? That would be very helpful.

How K plays into my proof is rather obvious, imo. It comes from the correspondence theorem, one of the most beautiful, helpful and basic of all basic group theory theorems: Correspondence Theorem - ProofWiki

Tonio

7. I guess we have not got to corresponedence theroem yet in class, howver the proof that you gave me for it makes perfect sense.
Thanks a lot.