(New page: Show that the direct product of <math>Z_2</math>, <math>Z_2</math>, <math>Z_2</math> has 7 subgroups of order 2. ---- This is fairly simple if you were to write out the subgroups. We hav...) |
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This is fairly simple if you were to write out the subgroups. | This is fairly simple if you were to write out the subgroups. | ||
We have: | We have: | ||
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(0,0,0) <-- has order 1 because it is the identity | (0,0,0) <-- has order 1 because it is the identity | ||
Revision as of 11:22, 23 February 2009
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)
