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

**rualin** · September 18th 2010, 07:13 PM

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 $D_n (n\geq 3)$ .

**rualin** · September 18th 2010, 07:15 PM

Ok... so my title is wrong... it's that union e...

**undefined** · September 18th 2010, 07:26 PM

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.

**rualin** · September 18th 2010, 07:31 PM

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.

**undefined** · September 18th 2010, 07:41 PM

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

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