Groups Identity element

[mikael27]

Homework Statement

Which of following statements are TRUE or FALSE. Why?

In any group G with identity element e

a) for any x in G, if x² = e then x = e.
b) for any x in G, if x² = x then x = e.
c) for any x in G there exists y in G such that x = y².
d) for any x, y in G there exists z in G such that y = xz.

Homework Equations

The Attempt at a Solution

for a) False
b) True

What reason i have to give for these answers?

 

[Dick]

Homework Statement

Which of following statements are TRUE or FALSE. Why?

In any group G with identity element e

a) for any x in G, if x² = e then x = e.
b) for any x in G, if x² = x then x = e.
c) for any x in G there exists y in G such that x = y².
d) for any x, y in G there exists z in G such that y = xz.

Homework Equations

The Attempt at a Solution

for a) False
b) True

What reason i have to give for these answers?

You gave answers for a) and b). You must have some reason for thinking they are right. What are those reasons?

 

[mikael27]

sorry a) True because x*x=e*e=e
b) False x*x=x but we can not conclude that x=e for a group

 

[Dick]

sorry a) True because x*x=e*e=e
b) False x*x=x but we can not conclude that x=e for a group

Oh, heck. Now you changed the right answers into wrong answers for no good reason. For a), list all the groups you know. For b) what happens if you multiply by x^(-1)?

 

[Deveno]

sorry a) True because x*x=e*e=e
b) False x*x=x but we can not conclude that x=e for a group

you are given that x² = e.

while it IS true that e² = e as well, it is NOT generally true that:

a² = b² implies a = b.

i suggest you re-consider your answer for (a), in light of groups of order 2.

if G = {e,x}, what must x^-1 be?

(hint: what happens if we say that x^-1 = e, so that x*e = __?)

b) why not? note that the problem is not claiming x² = x for EVERY element x of G, just for a particular one.

suppose x*x = x

multiply both sides (on the right, for example) by x^-1. what happens?

 

[mikael27]

a) False if G={e,x} the group is

e x
x e

if we have x^-1 = e
then x*x^-1=x*e
so e=x*eb) True

x^2=x
x*x=x
x*x*x^-1=x*x^-1
x*e=e

are these enough reasons ?

 

[Dick]

a) False if G={e,x} the group is

e x
x e

if we have x^-1 = e
then x*x^-1=x*e
so e=x*eb) True

x^2=x
x*x=x
x*x*x^-1=x*x^-1
x*e=e

are these enough reasons ?

It's good for b). For a) you seem to be picking the specific group Z_2. Which is a good example but the rest of what you say is garbage. x^(-1)=e makes no sense. Why do you think it does? x^(-1)=x, doesn't it?

 

[mikael27]

a) False if G={e,x} the group is

e x
x e

also

x*x^-1=e and that x*e=x=e*a

therefore x can not be equal to e

 

[Dick]

a) False if G={e,x} the group is

e x
x e

also

x*x^-1=e and that x*e=x=e*a

therefore x can not be equal to e

You picked a group where x is not equal to e, but where x^2=e. Isn't that enough to show what you want? That a) is False?

 

[mikael27]

Yes. so by showing this group we can conclude that there exist a group where x is not equal to e and hence this statement is false ?

 

[Dick]

Yes. so by showing this group we can conclude that there exist a group where x is not equal to e and hence this statement is false ?

Yes, exactly.

 

[mikael27]

Thank you. What about c and d. I think that c if false and that d is true. Do i have to use the group axioms here?

 

[Deveno]

Thank you. What about c and d. I think that c if false and that d is true. Do i have to use the group axioms here?

in general you want to do one of two things:

a) if it is true, PROVE it.

b) if it is not true, find a counter-example.

ok, so when you start out, you may have no idea if it's true or not. so you play with a few example groups, to try to see if it "might" be true.

what groups do you know about? i bet you already know that the integers under addition is a group, hmm?

so look at (c) with the integers in mind. note that since the operation is "+", that x*x isn't written x² anymore, because it's x+x, which is usually written 2x.

so in the integers, (c) would mean:

for every integer x, there is another integer y with x = 2y. do you think that is a true statement?

you could look at (d) with the integers in mind, as well. there, (d) says:

for every pair of integers, x and y, there is another integer z with y = x + z.

if this is true, what would z have to be?

now suppose that you look at (d) in light of the group of non-zero real numbers under multiplication, so it says:

for every x,y ≠ 0, there is z ≠ 0 with y = xz. can you "solve for z" in this case?

does this generalize to other groups? why, or why not?

 

[mikael27]

Acccording to the the above.

for c) x=y^2 so when we have addition operation "+" is going to be x=y+y=2y

for example the group (Z5, +) is :

0 1 2 3 4
1 2 3 4 0
2 3 4 0 1
3 4 0 1 2
4 0 1 3 4

If we choose random elements of the group we have:

Let x=2 and y=1 then x=2y is satisfied
but if we say that x=2 and y= 3 then x=2y is not satisfied

So it depends on the choice of x and y in G

Therefore the statement is false.

 

[mikael27]

d) y=xz so when we have addition operation "+" is going to be y=x+z

This statement is true only when z can be any real number so that y=x+z will be satisfied.

For the case of multiplication y=xz

for every x,y ≠ 0, there is z ≠ 0 with y = xz.

to find z :

y = xz
x^-1*y=x^-1*x*z
so z=x^-1*y

 

[Dick]

d) y=xz so when we have addition operation "+" is going to be y=x+z

This statement is true only when z can be any real number so that y=x+z will be satisfied.

For the case of multiplication y=xz

for every x,y ≠ 0, there is z ≠ 0 with y = xz.

to find z :

y = xz
x^-1*y=x^-1*x*z
so z=x^-1*y

That's it. You don't need to make a special mention of the case where '*' represents addition. In additive notation that looks like:

y=x+z
(-x)+y=(-x)+x+z=0+z
so z=(-x)+y.

They are the same thing. Just pick '*' or '+' for your group operation and stick with it.

 

[Dick]

Acccording to the the above.

for c) x=y^2 so when we have addition operation "+" is going to be x=y+y=2y

for example the group (Z5, +) is :

0 1 2 3 4
1 2 3 4 0
2 3 4 0 1
3 4 0 1 2
4 0 1 3 4

If we choose random elements of the group we have:

Let x=2 and y=1 then x=2y is satisfied
but if we say that x=2 and y= 3 then x=2y is not satisfied

So it depends on the choice of x and y in G

Therefore the statement is false.

No, you don't need to have x=2y for ANY two elements of the group. You need to show that for any x there exists a y such that x=2y. 0=0+0, 1=3+3, 2=1+1, 3=4+4, 4=2+2. This group isn't going to work as a counterexample. There is a solution to x=2y for any x.

 

[mikael27]

For c) This statement is true

x=y^2 so when we have addition operation "+" is going to be x=y+y=2y

so x=2y is always true as there is a number y to satisfy this expression.

So

d) This statement is true

y=xz when we have addition operation "+" is going to be

y=x+z
(-x)+y=(-x)+x+z=0+z
so z=(-x)+y.

which is true

 

[Dick]

For c) This statement is true

x=y^2 so when we have addition operation "+" is going to be x=y+y=2y

so x=2y is always true as there is a number y to satisfy this expression.

So

d) This statement is true

y=xz when we have addition operation "+" is going to be

y=x+z
(-x)+y=(-x)+x+z=0+z
so z=(-x)+y.

which is true

c) is true for Z5. It IS NOT TRUE for every group. Try to think of a group where it's not true. And you didn't have to switch your group notation from multiplicative to additive. Just stick with multiplicative, since the problem is presented that way.

 

[mikael27]

ok so we have:

For c) This statement is false

counterexample group Z2.

0 1
1 0

if x=0 and y=1

then x is not equal tp y^2

therefore the statement is false

for d) This statement is true

y=xz

for every x,y ≠ 0, there is z ≠ 0 with y = xz.

to find z :

y = xz
x^-1*y=x^-1*x*z
so z=x^-1*y

which is true

 

[Dick]

ok so we have:

For c) This statement is false

counterexample group Z2.

0 1
1 0

if x=0 and y=1

then x is not equal tp y^2

therefore the statement is false

for d) This statement is true

y=xz

for every x,y ≠ 0, there is z ≠ 0 with y = xz.

to find z :

y = xz
x^-1*y=x^-1*x*z
so z=x^-1*y

which is true

For c) you are doing exactly what you did before. Reread post 17. At least you've got a better group to find a counterexample in. Your group table says 0*0=0, 0*1=1*0=1, 1*1=0. If x=0, is there a y such that y^2=0? What about if x=1? Any element satisfying y^2=1? For d) where is this "≠ 0" stuff coming from? In a group, EVERY element has an inverse.

 

[mikael27]

For c) This statement is false

counterexample group Z2.

0 1
1 0

From the group table we have : 0*0=0, 0*1=1*0=1, 1*1=0
which can be separated to two cases

1) 0*0=1*1=0 x=0 , there is an element which satisfy y^2=0.
2) 0*1=1*0=1 x=1 , there is no any element which satisfy y^2=1

so y^2 is not equal to x

This statement is false

for d) This statement is true

y=xz

to find z :

y = xz
x^-1*y=x^-1*x*z
so z=x^-1*y

which is true

 

[Dick]

For c) This statement is false

counterexample group Z2.

0 1
1 0

From the group table we have : 0*0=0, 0*1=1*0=1, 1*1=0
which can be separated to two cases

1) 0*0=1*1=0 x=0 , there is an element which satisfy y^2=0.
2) 0*1=1*0=1 x=1 , there is no any element which satisfy y^2=1

so y^2 is not equal to x

This statement is false

for d) This statement is true

y=xz

to find z :

y = xz
x^-1*y=x^-1*x*z
so z=x^-1*y

which is true

That's good enough. Now are you sure you understand it and aren't just pasting peoples suggestions together?

 

[mikael27]

yes am sure about it. thank you

 

[Deveno]

For c) This statement is false

counterexample group Z2.

0 1
1 0

From the group table we have : 0*0=0, 0*1=1*0=1, 1*1=0
which can be separated to two cases

1) 0*0=1*1=0 x=0 , there is an element which satisfy y^2=0.
2) 0*1=1*0=1 x=1 , there is no any element which satisfy y^2=1

so y^2 is not equal to x

This statement is false

it's clear to me you don't even understand what you wrote here. i think you find this all very confusing. please understand that if you turn in the right answer, but with the wrong reasoning, you aren't likely to get any credit for it. the "why" of an answer is much more important than the answer itself.

for d) This statement is true

y=xz

to find z :

y = xz
x^-1*y=x^-1*x*z
so z=x^-1*y

which is true

why is this true? at this point, almost every step you take should have a reference to one of the group axioms, justifying it. because groups don't act like the numbers you are used to.