Generators of Z6


by ma3088
Tags: abstarct algebra, generators
ma3088
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#1
Feb4-13, 03:41 PM
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I understand how {1} and {5} are generators of Z6.

{1} = {1, 2, 3, 4, 5, 0} = {0, 1, 2, 3, 4, 5}
{5} = {5, 4, 3, 2, 1, 0} = {0, 1, 2, 3, 4, 5}

But my book also says that {2, 3} also generates Z6 since 2 + 3 = 5 such as {2,3,4} and {3,4} I believe. Thus every subgroup containing 2 and 3 must also be 5 except for {2,4}.

Can someone explain this to me? Ty in advance.
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jbunniii
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Feb4-13, 04:09 PM
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Your notation is incorrect. {1} is a set containing one element, 1. {1, 2, 3, 4, 5, 0} is a set containing six elements. Therefore {1} does not equal {1, 2, 3, 4, 5, 0}.

The usual notation for the group generated by a set is a pair of angle brackets: <{1}> denotes the group generated by the set {1}. It is true that <{1}> = <{5}> = Z6. It is also (trivially) true that <{1, 2, 3, 4, 5, 0}> = Z6.

Note that in general, if S is a subset of Z6, <S> is the smallest subgroup of Z6 which contains all of the elements of S. If S is a subgroup, then S = <S>. Also, it's easy to verify that if S [itex]\subseteq[/itex] T, then <S> [itex]\subseteq[/itex] <T>.

Now what about <{2,3}>? This is a group, by definition, so it must be closed under addition. Thus <{2,3}> must contain 5 because 2+3=5. In other words, {5} [itex]\subseteq[/itex] <{2,3}>. Therefore Z6 = <{5}> [itex]\subseteq[/itex] <{2,3}>. For the reverse containment, we have {2,3} [itex]\subseteq[/itex] {1,2,3,4,5,0}, so <{2,3}> [itex]\subseteq[/itex] <{1,2,3,4,5,0}> = Z6. We conclude that <{2,3}> = Z6.


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