# Homework Help: Groups Identity element

1. Feb 21, 2012

### mikael27

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution

for a) False
b) True

What reason i have to give for these answers?

2. Feb 21, 2012

### Dick

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

3. Feb 21, 2012

### 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

4. Feb 21, 2012

### Dick

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)?

5. Feb 21, 2012

### Deveno

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?

6. Feb 21, 2012

### 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*e

b) True

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

are these enough reasons ?

7. Feb 21, 2012

### Dick

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?

8. Feb 21, 2012

### 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

9. Feb 21, 2012

### Dick

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?

10. Feb 22, 2012

### 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 ?

11. Feb 22, 2012

### Dick

Yes, exactly.

12. Feb 22, 2012

### 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?

13. Feb 22, 2012

### Deveno

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?

14. Feb 22, 2012

### 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.

15. Feb 22, 2012

### 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

16. Feb 22, 2012

### Dick

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.

17. Feb 22, 2012

### Dick

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.

18. Feb 22, 2012

### 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

19. Feb 22, 2012

### Dick

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.

20. Feb 22, 2012

### 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