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Show that the direct product of <math>Z_2</math>, <math>Z_2</math>, <math>Z_2</math> has 7 subgroups of order 2. | Show that the direct product of <math>Z_2</math>, <math>Z_2</math>, <math>Z_2</math> has 7 subgroups of order 2. | ||
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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. | 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. | ||
--[[User:Podarcze|Podarcze]] 16:22, 23 February 2009 (UTC) | --[[User:Podarcze|Podarcze]] 16:22, 23 February 2009 (UTC) | ||
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| + | Another way of looking at the problem is that there are a total of 8 subgroups because there are two options for each of three seats => 2^3 = 8. Z2 has elements {0,1} where |0| = 1 and |1| = 2. Thus any subgroup with a 1 in any of the three seats will have an order of 2. This happens in all of the subgroups except one. Namely the identity (0,0,0). Therefore the total number of subgroups with order 2 is 8-1 = 7. | ||
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| + | -K. Brumbaugh | ||
Latest revision as of 14:17, 24 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)
Another way of looking at the problem is that there are a total of 8 subgroups because there are two options for each of three seats => 2^3 = 8. Z2 has elements {0,1} where |0| = 1 and |1| = 2. Thus any subgroup with a 1 in any of the three seats will have an order of 2. This happens in all of the subgroups except one. Namely the identity (0,0,0). Therefore the total number of subgroups with order 2 is 8-1 = 7.
-K. Brumbaugh
