My Math Forum Subgroups of an arbitrary groups with square commutation.

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January 14th, 2013, 04:28 AM   #1
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Subgroups of an arbitrary groups with square commutation.

Exercise:
Let G be a group. Prove the following:
Let$C' = \{a \in G : (ax){^2} = (xa){^2} \quad \forall x \in G\}$. 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:
$(ex)^2= (xe)^2$
$x^2= x^2$
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.
$(ax)^2= (xa)^2 \Longleftrightarrow axax = xaxa$
multiplying to the left by $a^{-1}$
$a^{-1}axax= a^{-1}xaxa$
$xax= a^{-1}xaxa \quad \forall x \in G$
multiplying to the right by $a^{-1}$
$xaxa^{-1}= a^{-1}xaxaa^{-1}$
$xaxa^{-1}= a^{-1}xax \quad \forall x \in G$

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) $(ax)^2= (xa)^2 \Longleftrightarrow axax = xaxa$
ii) $(bx)^2= (xb)^2 \Longleftrightarrow bxbx = xbxb$
I) multiplying to the left in i) we obtain:
$baxax= bxaxa \quad \forall x in G$

I don't know if this is the right way to proceed, can you tell me some suggestions please? many thanks!

January 17th, 2013, 03:18 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 $(a^{-1}x)^{2}= (xa^{-1})^{2}$ ?
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:
$(ax)^2= (xa)^2 \wedge (bx)^2 = (xb)^2 \Rightarrow (abx)^2 = (xab)^2 \vee (bax)^2 = (xba)^2$
but the problems still remain the same.

 January 20th, 2013, 01:08 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, 03:33 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 $(a^n)^m= a^{nm}$?
so in the case of the exercise it should be $(g^{-1})^2= g^{(-1)(2)} = g^{-2}$ ?

Quote:
 Originally Posted by Deveno $(a^{-1})g^2= [((g^{-1})^2)a]^{-1}$
i don't understand why you have put -1 outside the square parenthesis. if i replace $g^2$ with $(g^{-1})^2$, why not obtain $(g^{-1})^2a^{-1}$ without square parentheis?

 January 21st, 2013, 08:13 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.
January 22nd, 2013, 03:46 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 $(a^{-1})^k$ the inverse of $a^k$?
so, if I'm right, all the procedure would be as:
$(a^k)(a^{-1})^k= e$
$(a^{-1})^k= (a^k)^{-1}$
$(a^k)^{-1}= (a^k)^{-1}$ <-- here in the first member i swap the inner exponent with the outer one. is it correct?
$(a^{-k})^1= (a^k)^{-1}$
$a^{-k}a^k= (a^k)^{-1}(a^k)$
$a^{-k}a^k= e$
$a^{-k+k}= e$
$a^0= e$
$e= e$

great! that's ok!! all your remaining suggestions has been very clear and coincise! thank you very much!!

January 31st, 2013, 06:24 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 $(a^{-1}x)^{2}= (xa^{-1})^{2}$ ?
if the answer is yes, then my development of this point is also wrong!
Let $a\in C'$ and $x\in G$. The trick is to start with $(ax^{-1})^2=(x^{-1}a)^2$ (since $(ax)^2=(xa)^2$ for all $x\in G$). After expanding and some juggling, you should arrive at $(a^{-1}x)^2=(xa^{-1})^2$.

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:
$(ax)^2= (xa)^2 \wedge (bx)^2 = (xb)^2 \Rightarrow (abx)^2 = (xab)^2 \vee (bax)^2 = (xba)^2$
but the problems still remain the same.
Let $a,b\in C'$ and $x\in G$. Then

$\begin{array}{ll}
{} & ([ab]x)^2\\
= & (a[bx])^2\\
= & ([bx]a)^2\\
= & (b[xa])^2\\
= & ([xa]b)^2\\
= & (x[ab])^2\end{array}$

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