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javedansari
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If x and y be elements in a group G such that xy ε Z(G)
prove that xy = yx..
prove that xy = yx..
javedansari said:If x and y be elements in a group G such that xy ε Z(G)
prove that xy = yx..
Tenshou said:So you are trying to prove commutativity of a group? Well... Let
##xy \in G## since it must be a obey closure of a group. now, since the product of two elements are in a group let there exist an element ##e## such that ## xy= e##, since we are assuming the existing of an element ##e## we should show that this object is "unique" so we set ## yc=e## where ##c \in G## we can now show that, ##xy=yxe## this is clearly true since ##\forall x \in G[xe=x]## is true, clearly ##xy=yx(yc)## using a little bit associativity and hypothesis we get ##xy=y(xy)c \implies xy=yc = e## thus establishing that ##\forall x,y \in G [xy=e \implies yx=e]## since those elements are the same,do the math .
Here is a better one
suppose that the a member ##x \in G## such that ##xx=e## is true, then ##xe=ex=x## furthermore we can show that ##x(xx)=(xx)x=x \implies (xx)x =x(xx) =x## by substitution and associativity it becomes ## ex=xe=x## much easier.
So, if xy doesn't commute with x inverse what would happen to Z(G) would this product not be in the center?lavinia said:The center of the group is those elements that commute with everything.
In particular, xy must commute with x^-1
Not that x and y do not need to be in the center of the group. Can you construct an example?
micromass said:I would appreciate it if you wouldn't give answers on subjects you are not qualified in.
Tenshou said:So, if xy doesn't commute with x inverse what would happen to Z(G) would this product not be in the center?
lavinia said:Try to figure this out. It isn't difficult.
HallsofIvy said:No, in any group at least the group identity commutes with all elements. So the "center of the group" consists of at least the identity. In this case, xy is in the center because it is given as the hypothesis of the problem. Sorry, but I can't make heads or tails out of your post #2. You start by arguing that xy must be something, and call it "e". But then state [itex]\forall x \in G [xe= x][/itex], assuming that e is the identity. That makes no sense at all!
He just made the thread today :/ I wanted to solve this it didn't seem to difficult I mean I have done it before, just without having this proper(or not) subgroup in there. And I didn't understand what I was doing wrong, and when this question popped up I thought I would try it out and see if I can get some critiques on what I did wrong. And maybe what I could possibly do better.verty said:Tenshou: This was Javedansari's question. You have sort of railroaded the thread. I do that too sometimes but usually only when enough hints have been given or the poster has not replied for a couple of days.
And it's not like your posts are getting any clearer or more justified. My recommendation is to try work things out by yourself and if you get stuck, ask questions that are specific to what you wish to know without giving too much away.
But actually, a much stronger recommendation from me is to study your school math instead. Focusing on this higher level stuff is a mistake when we know that there are things more important to know that you must learn.
Now, can we wait for Javedansari to respond? Thanks.
Tenshou said:He just made the thread today :/ I wanted to solve this it didn't seem to difficult I mean I have done it before, just without having this proper(or not) subgroup in there. And I didn't understand what I was doing wrong, and when this question popped up I thought I would try it out and see if I can get some critiques on what I did wrong. And maybe what I could possibly do better.
Tenshou said:Sorry I am asking the wrong question. I mean since x and y don't need to be in the center of the group, then why does the product need to be in the center? I mean the group could be centerless, right?
micromass said:So please, do not hijack threads anymore, but rather create your own thread.
The Algebra theorem is a mathematical law or rule that proves the validity of algebraic equations and manipulations. It is the foundation of algebra and is used to solve complex equations and problems.
The main types of Algebra theorem include the distributive property, associative property, commutative property, and the inverse property. These theorems help to simplify and solve equations by manipulating the terms and operations.
The Algebra theorem is used in various real-life scenarios such as calculating distances, determining the value of investments, predicting future outcomes, and solving engineering problems. It is also used in fields such as physics, economics, and statistics.
One example of the Algebra theorem in action is the distributive property. For instance, when expanding the expression (x + y)(x + z), the distributive property states that the result will be x(x + z) + y(x + z). This allows us to simplify the equation and solve for x and y.
No, the Algebra theorem is only applicable to linear equations, which involve variables raised to the first power. It cannot be applied to equations with variables raised to a higher power, such as quadratic or exponential equations.