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- Dec 15th 2007, 01:05 AM #1

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- Dec 15th 2007, 01:17 AM #2

- Dec 15th 2007, 01:21 AM #3
as i see it from your example, the set is

and defines by multiplication of elements modulo 10..

am i right?

from that, which do you think takes the role of the identity? it is one of the subgroups..

also, if you want to have another subgroup, this must contain the identity..

now if you have the set containing the identity and add another element, which will you add that would make a group?

another, in that set, which elements are inverses?

*the reason i do this is simply because, the set is finite..

- Dec 15th 2007, 09:57 AM #4

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- Dec 15th 2007, 02:38 PM #5

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By Lagrange's theorem by possible subgroups can have orders: 1 or 2 or 4.

Note if the order is 1 then it is simply the trivial subgroup.

Note if the order is 4 then it is the improper (full) subgroup.

Now if the order is 2 one of the elements needs to be the identity element. Now form all possible 2 element subsets one of which contains the identity element and see if the sub-group properties are satisfied.*

*)Here is an even easier way. {e,a} is a subgroup if and only if it is closed under the binary operation. Since ee = e, and ea=ae=a, it means we just need to check whether a^2 is in {e,a}, i.e. a^2 = e.

- Dec 15th 2007, 08:06 PM #6
the theorem TPH gave would really help you a lot..

however, if you only know the definition of a subgroup of a group, then you could only rely on the axioms a group must satisfy..

the set is finite..

look at 3 axioms, one states that it must have an identity, and each element must have an inverse..

the order of the set is small (only 4), thus you can just eliminate all the subsets which does not form a group..

list all the subsets of singletons, pairwise and tiplewise.. then do the elimination process..

exmaple.. {2}, {4}, {6}, {8}.. among these singletons, only {6} forms a subgroup of the set, because, it is closed on *, it is associative, (in fact) it is the identity and thus its inverse is itself.. while the others are not since they are not closed..

try to do it pairwise, like {2,4},{2,6},... and eliminate those does not satisfy the axioms.. take note that 2 and 8 are inverses, thus, when the set contains 2, it must also contain 8.. therefore remove the subset that does not contain**both**2 and 8..

do it tripplewise.. note that TPH gave you Lagrange's Thm, thus you should get no subgroup of the group containing 3 elements..

i hope, this reply could help you..