# Thread: Lagranges Theorem question (Not sure if this is in the right section)

1. ## Lagranges Theorem question (Not sure if this is in the right section)

Let h be a subgroup of a group G of order 6. Show that if |H|=3 then every left coset of H is a right coset. Is this necessarily true if |H|=2?

I think this is a lagrane theorem question, I understand that they are asking me if H is a normal subgroup, but I am unsure how to prove this.
I have come across a proof in my notes saying that any subgroup of order 2 is normal, will this suffice for the second part of the question?
Thank you

2. Originally Posted by nlews
Let h be a subgroup of a group G of order 6. Show that if |H|=3 then every left coset of H is a right coset. Is this necessarily true if |H|=2?

I think this is a lagrane theorem question, I understand that they are asking me if H is a normal subgroup, but I am unsure how to prove this.
I have come across a proof in my notes saying that any subgroup of order 2 is normal, will this suffice for the second part of the question?
Thank you
Yes, this is the section of the forum you want!

I think you are getting a tad confused with order and index. The theorem from your notes should, I believe, be saying that every subgroup of index 2
is normal. A subgroup $H$ has index $n$ if $|G|/|H| = n$, where $|H|$ is, as I'm sure you know, the order (aka number of elements) of $H$.

Therefore, you have the first result, as the subgroup you are given has index two, and so is normal.

For the second result, which groups of order 6 do you know? There is one abelian one, but you can forget about this as every subgroup of an abelian group is normal. So, take your non-abelian group of order 6. Take an element of order 2 in it. Does this element generate a normal subgroup?

3. Oh yes it does say index! That makes so much more sense for part one. Thank you.

for part 2)
The non abelian one would be D3.

4. Originally Posted by nlews
Oh yes it does say index! That makes so much more sense for part one. Thank you.

for part 2)
The non abelian one would be D3.
Yup. So, take some element of order 2 (just take the generating element of order 2, call it $\beta$). Is the subgroup that it generates normal? Look at the element $\alpha^{-1}\beta\alpha$ where $\alpha$ is the generating element of order 3.