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January 14th, 2013, 03:28 AM | #1 | ||||
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7 | Subgroups of an arbitrary groups with square commutation.
Exercise: Let G be a group. Prove the following: Let I have done in this way: To demonstrate that C' is a subgroup of G, is important that: 1) C' is a non-empty set. 2) C' is a subset of G. 3) C' is closed with respect the inverses. 4) C' is closed with respect the product. Let's go: Quote:
From one axiom of group, for every element x in G, exists one identity element e such that xe=x=ex. Applying in the exercise we have: hence C' has at least one element a=e within. Quote:
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multiplying to the left by multiplying to the right by i'm trying similar for 4) but without success... Quote:
i) ii) I) multiplying to the left in i) we obtain: I don't know if this is the right way to proceed, can you tell me some suggestions please? many thanks! | ||||
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January 17th, 2013, 02:18 AM | #2 | ||
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7 | Re: Subgroups of an arbitrary groups with square commutation Quote:
the doubt increases. Do I have to demonstrate ![]() if the answer is yes, then my development of this point is also wrong! Quote:
but the problems still remain the same. ![]() ![]() | ||
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January 20th, 2013, 12:08 AM | #3 |
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88 | Re: Subgroups of an arbitrary groups with square commutation
there is a "trick" to this: suppose (ag)^2 = (ga)^2 for ALL g in G. well ga^-1 is certainly an element of G, so: (a(ga^-1))^2 = ((ga^-1)a)^2 (you can, i hope, see that the elements ga^-1 run through all the elements of G as g runs through all the elements of G), for any g in G. expanding this, we have: (aga^-1)(aga^-1) = g^2 that is: ag^2a^-1 = g^2, equivalently: ag^2 = g^2a. so "square commutation" is the same condition as "commutes with all squares". the latter condition is LOTS easier to deal with. now if ag^2 = g^2a, for all g in G, and the same is true of b, then: (ab)g^2 = a(bg^2) = a(g^2b) = (ag^2)b = (g^2a)b = g^2(ab), so we have closure. now if a commutes with g^2 for ANY g, this means a commutes with (g^-1)^2, as well. so: a(g^-1)^2 = ((g^-1)^2)a but (a^-1)g^2 = [((g^-1)^2)a]^-1 = [a(g^-1)^2]^-1 = g^2(a^-1), so a^-1 commutes with all g^2 in G if a does. |
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January 21st, 2013, 02:33 AM | #4 | |||
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7 | Re: Subgroups of an arbitrary groups with square commutation
thank you very much Deveno !! but I have some doubts yet. Quote:
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so in the case of the exercise it should be Quote:
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January 21st, 2013, 07:13 PM | #5 |
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88 | Re: Subgroups of an arbitrary groups with square commutation
yes, in groups for any integers k and m we have: (a^k)^m = a^(km). this is because a^(-k) is DEFINED to be (a^-1)^k = (a^k)^-1, so that: (a^(-k))(a^k) = a^(-k+k) = a^0 = e. the reason why i write: (a^-1)g^2 = [(g^-1)^2a]^-1 is because we don't know yet (at this stage) that a^-1 commutes with any square, but we DO know that a does. so i want an expression involving a, not a^-1. convince yourself that x = y if and only if x^-1 = y^-1. since we are dealing with a perhaps non-abelian group we CANNOT assume: (xy)^n = (x^n)(y^n). in general this is not true. however, it IS true for any group that (xy)^-1 = y^-1x^-1 <---this is what i'm using to get an expression involving a instead of a^-1. let me run it by you slower: (a^-1)g^2 = [((a^-1)g^2)^-1]^-1 (for any x in G, (x^-1)^-1 = x) = [(g^2)^-1(a^-1)^-1]^-1 (working on the inner inverse, and using (xy)^-1 = y^-1x^-1, with x = a^-1, and y = g^2) = [((g^2)^-1)a]^-1 (using the fact that (a^-1)^-1 = a) = [((g^-1)^2)a]^-1 (using the fact that (g^2)^-1 is (g^-1)^2, look: (g^2)(g^-1)^2 = (gg)(g^-1g^-1) = g(gg^-1)g^-1 = geg^-1 = gg^-1 = e, so (g^-1)^2 must be the (unique!!!!) inverse of g^2). = [a(g^-1)^2]^-1 (still working "inside the square brackets" using the fact that a commutes with ANY square, including the square of g^-1) = ([(g^-1)^2]^-1)(a^-1) (using the inverse rule, this time with x = a, and y = (g^-1)^2) = ([(g^2)^-1]^-1)(a^-1) (we did this above: the inverse of g^2 IS the square of g^-1) = (g^2)(a^-1) (again using that (x^-1)^-1 = x)...which shows that a^-1 commutes with any square if a does. |
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January 22nd, 2013, 02:46 AM | #6 | |
Senior Member Joined: Jan 2013 From: Italy Posts: 154 Thanks: 7 | Re: Subgroups of an arbitrary groups with square commutation Quote:
so, if I'm right, all the procedure would be as: great! that's ok!! all your remaining suggestions has been very clear and coincise! thank you very much!! ![]() | |
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January 31st, 2013, 05:24 PM | #7 | ||||
Member Joined: Jan 2013 Posts: 93 Thanks: 0 | Re: Subgroups of an arbitrary groups with square commutation Quote:
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