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January 14th, 2013, 03:28 AM   #1
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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 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:
1) C' is a non-empty set.
For every element x in G, what is the element a in G which satisfies the equation given in the exercise?
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:
2) C' is a subset of G.
because e in G.

Quote:
3) C' is closed with respect the inverses.

multiplying to the left by


multiplying to the right by




i'm trying similar for 4) but without success...

Quote:
4) C' is closed with respect the product.
we want to show that if the elements a,b are in C', then also the element ab is in C'.
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
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Re: Subgroups of an arbitrary groups with square commutation

Quote:
Originally Posted by beesee
3) C' is closed with respect the inverses.
we want to show that if a is in C' then a^-1 also is in C'.
the doubt increases. Do I have to demonstrate ?
if the answer is yes, then my development of this point is also wrong!

Quote:
Originally Posted by beesee
4) C' is closed with respect the product.
for more clarification i have set up this point in this way:

but the problems still remain the same.
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January 20th, 2013, 12:08 AM   #3
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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
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Re: Subgroups of an arbitrary groups with square commutation

thank you very much Deveno !! but I have some doubts yet.

Quote:
Originally Posted by Deveno
(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.
ok. briefly you have used the fact that aa^-1 = e. so you rewrite g in a different way.

Quote:
Originally Posted by Deveno
now if a commutes with g^2 for ANY g, this means a commutes with (g^-1)^2
Pardon. I have a doubt yet! here in group theory, is the rule to follow the same of the power of power ?
so in the case of the exercise it should be ?

Quote:
Originally Posted by Deveno
i don't understand why you have put -1 outside the square parenthesis. if i replace with , why not obtain without square parentheis?
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January 21st, 2013, 07:13 PM   #5
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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
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Re: Subgroups of an arbitrary groups with square commutation

Quote:
Originally Posted by Deveno
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.
so the question is: is the inverse of ?
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!!
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January 31st, 2013, 05:24 PM   #7
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Re: Subgroups of an arbitrary groups with square commutation

Quote:
Originally Posted by beesee
Quote:
Originally Posted by beesee
3) C' is closed with respect the inverses.
we want to show that if a is in C' then a^-1 also is in C'.
the doubt increases. Do I have to demonstrate ?
if the answer is yes, then my development of this point is also wrong!
Let and . The trick is to start with (since for all ). After expanding and some juggling, you should arrive at .


Quote:
Originally Posted by beesee
Quote:
Originally Posted by beesee
4) C' is closed with respect the product.
for more clarification i have set up this point in this way:

but the problems still remain the same.
Let and . Then

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