# D8 subgroups

• Oct 8th 2011, 04:12 PM
dwsmith
D8 subgroups
Easy question. How do I find the subgroups of \$\displaystyle D_8\$?

\$\displaystyle D_8=<r,s:r^4=s^2=1, \ rs=sr^{-1}>\$

\$\displaystyle D_8=\{1,r,r^2,r^3,s,sr,sr^2,sr^3\}\$
• Oct 8th 2011, 04:42 PM
Bernhard
Re: D8 subgroups
See Deveno's posts in answer to my post "Normal Subgroups in D4 "

Peter
• Oct 8th 2011, 04:52 PM
dwsmith
Re: D8 subgroups
Quote:

Originally Posted by Bernhard
See Deveno's posts in answer to my post "Normal Subgroups in D4 "

Peter

So the Centralizer of each element is a subgroup then?

And normal subgroups are in the center?

• Oct 8th 2011, 04:57 PM
Bernhard
Re: D8 subgroups
Sorry.

I was too quick in referrring you to my post - I was looking for normal subgroups only.

Peter
• Oct 8th 2011, 05:07 PM
dwsmith
Re: D8 subgroups
Quote:

Originally Posted by Bernhard
Sorry.

I was too quick in referrring you to my post - I was looking for normal subgroups only.

Peter

It didn't matter I learned something from it.

So the cyclic subgroups are \$\displaystyle <r>=<r^3>\$, correct?
• Oct 8th 2011, 06:03 PM
Bernhard
Re: D8 subgroups
Yes, but I think there may be others of order 2 such as <s> = {s, e} and <sr> = {sr, e}

Do you agree?

Peter
• Oct 8th 2011, 06:13 PM
dwsmith
Re: D8 subgroups
Quote:

Originally Posted by Bernhard
Yes, but I think there may be others of order 2 such as <s> = {s, e} and <sr> = {sr, e}

Do you agree?

Peter

How were you able to identify those at cyclic groups?
• Oct 8th 2011, 06:36 PM
Bernhard
Re: D8 subgroups
Just by checking the subgroups generated by the elements concerned - no smart method - just working through the elements of the group checking the subgroups generated by each element. If the group concerned was very much bigger you would have to be more analytic I guess.

Another possibility is <\$\displaystyle r^2 \$> = {e, \$\displaystyle r^2 \$} and of course there is {e}.

Peter