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Sekky

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Registered: 2007-01-12

Posts: 181

Group Theory

http://www.maths.leeds.ac.uk/~marsh/MAT … 607/q4.pdf

I'm struggling with qs 3,4 and 7 on this sheet, any pointers?

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Group Theory

#3.

EDIT: Making proof a bit simpler …

I’ll work on #4 and #7 later.

Last edited by JaneFairfax (2007-03-15 09:16:46)

Sekky

Member

Registered: 2007-01-12

Posts: 181

Re: Group Theory

#3.

EDIT: Making proof a bit simpler …

I’ll work on #4 and #7 later.

Ok, for the third line, how does that imply that it is proper, couldn't a theoretically generate the entire group?

and for line four, why does a^m^n = e?

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Group Theory

Ok, for the third line, how does that imply that it is proper, couldn't a theoretically generate the entire group?

In short, no.  If a^m = e, then it must be that the order of <a> divides m.  This is fairly easily provable, but typically most will let you just state it as fact as it is a very common theorem.  Since the order of <a> divides m, and m < mn, it must be that the order of <a> is less than mn, i.e. the size of the group.

and for line four, why does a^m^n = e?

For all elements a of a group G, a^|G| = e.  This is a corollary to Lagrange's theorem.  As |G| = mn, a^mn = a^m^n = e.

For 4, you can start by eliminating any subset such that the size of the subset does not divide the size of the group.  Z8 is cyclic, so all subgroups of Z8 must be cyclic.  So all you need to do is compute:

<1>, <2>, <3>, ..., <7>

And find which of these are unique.

7.  If there is an element a of Zp* that is of order two, then a^2 = 1 (mod p).  In other words, p | (a^2 - 1) = (a + 1)(a - 1).  So either p | (a + 1) or p | (a - 1).  If p | (a+1), then a = -1 = p-1 (mod p).  If p | (a - 1), then a = 1 (mod p).

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Group Theory

#4 looks easy; it might just involve a tedious listing out of subgroups. I’ll do #7 first as that looks more interesting.

#7

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Group Theory

If a^m = e, then it must be that the order of <a> divides m.  This is fairly easily provable, but typically most will let you just state it as fact as it is a very common theorem.  Since the order of <a> divides m, and m < mn, it must be that the order of <a> is less than mn, i.e. the size of the group.

If a[sup]m[/sup] = e, it is enough to know that the order of <a> is less than or equal to m. That is enough to answer this particular question.

(And of course since a is not the identity, the order of <a> must be greater than 1.)

Last edited by JaneFairfax (2007-03-15 10:06:30)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Group Theory

#4 (There’s a part (a) but there doesn’t seem to be a part (b) for this question.)

I was right. The question involves nothing more than making lists of subsets of

.

Last edited by JaneFairfax (2007-03-15 12:14:41)