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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. Prove that C' is a subgroup of G. I have done in this way: To demonstrate that C' is a subgroup of G, is important that: 1) C' is a nonempty 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!  
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.  
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. 
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:
 
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 nonabelian 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. 
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: < here in the first member i swap the inner exponent with the outer one. is it correct? great! that's ok!! all your remaining suggestions has been very clear and coincise! thank you very much!!  
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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