# Subset of all order 2 elements of D_n not a subgroup of D_n

• Sep 18th 2010, 07:13 PM
rualin
Subset of all order 2 elements of D_n not a subgroup of D_n
For some reason I'm having trouble with this simple exercise. I'm sure I'm simply overlooking something. Thanks.

Show that $\{x \in D_n | x^2=e\}$ is not a subgroup of $D_n (n\geq 3)$.
• Sep 18th 2010, 07:15 PM
rualin
Ok... so my title is wrong... it's that union e...
• Sep 18th 2010, 07:26 PM
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Quote:

Originally Posted by rualin
For some reason I'm having trouble with this simple exercise. I'm sure I'm simply overlooking something. Thanks.

Show that $\{x \in D_n | x^2=e\}$ is not a subgroup of $D_n (n\geq 3)$.

I believe that set is precisely the reflections plus the identity, plus the 180 degree rotation if n is even, and closure does not hold.
• Sep 18th 2010, 07:31 PM
rualin
It's reflections, the identity, and if n is even, the rotation by 180 degrees. It is obvious to me that the subset is not a subgroup because it is not closed but I just can't seem to be able to prove it formally.
• Sep 18th 2010, 07:41 PM
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Quote:

Originally Posted by rualin
It's reflections, the identity, and if n is even, the rotation by 180 degrees. It is obvious to me that the subset is not a subgroup because it is not closed but I just can't seem to be able to prove it formally.

The composition of two reflections is a rotation. When you perform one reflection, the only way to get back to the original orientation via reflection is to do the same reflection again. Thus, since you have at least two (actually 3) reflections in your set, choosing any two of them will result in a rotation that is not the identity, and thus not in the set.

Edit: Sorry forgot about the 180 degree rotation again... I'm rethinking

Edit 2: Ah we can just use that $S_iS_j=R_{i-j}$

http://en.wikipedia.org/wiki/Dihedra...roup_structure