Show that the direct product of $ Z_2 $, $ Z_2 $, $ Z_2 $ has 7 subgroups of order 2.
This is fairly simple if you were to write out the subgroups. We have: (0,0,0) <-- has order 1 because it is the identity
(0,0,1), (0,1,0), (0,1,1), (1,0,0), (1,0,1), (1,1,0), (1,1,1),
The remaining 7 have an order of two because if you were to add because this set is under addition, you would get the identity for all of these subgroups. --Podarcze 16:22, 23 February 2009 (UTC)
