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### Proving an Element is in a Group

```Date: 04/12/2002 at 21:38:10
From: Jill
Subject: Proving an Element in a Group

Hi Dr. Math,

My friend and I are having difficulty in proving this problem. Can you

If G is a group, prove that the only element g in G with g^2 = g is 1.

Thanks so much,
Jill and Tonya
```

```
Date: 04/12/2002 at 23:43:09
From: Doctor Paul
Subject: Re: Proving an Element in a Group

Clearly, the identity satisfied g^2 = g. Suppose by way of
contradiction that there is another element of G, call it x, such
that x^2 = x.  Then in particular, x is not equal to the identity.
Since x is an element of a group, x^(-1) is in G as well.

Then multiplying both sides of x^2 = x by x^(-1) gives

Thus the desired result holds.

I hope this helps.  Please write back if you'd like to talk about
this some more.

- Doctor Paul, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 04/13/2002 at 11:10:23
From: Jill
Subject: Proving an Element in a Group

Dr. Paul,

Thank you so much for your help. We have a follow-up question for you.
We are having trouble understanding the whole idea of the identity.
Do we have to find what the identity is, or do we just know it?  We
have had proofs talking about the identity before and it confuses us.
```

```
Date: 04/13/2002 at 12:35:42
From: Doctor Paul
Subject: Re: Proving an Element in a Group

The identity elements depend on the group G.

Recall that a group is a set together with a binary association, often
denoted *, that satisfies:

1. the binary operation is associative
2. there is an identity element, denoted e, that satisfies:
e*x = x*e = x for all x in G.
3. every element is invertible.  That is, for all x in G,
there exists an element y in G such that x*y = e.

Of course, * denotes a generic binary operation. In practice, we
usually know what the operation is and so we use the appropriate

Let's look at the group of integers together with ordinary addition,
often denoted (Z, +). We want to find the identity element in this
group. That is, we want to find an element of the group (i.e., an
integer) such that when any other element of the group is added to
this particular element, we get the "other" element back.

That is, we want to find the element e such that

e + x = x for all integers x.

I think it's pretty clear that this forces e = 0.

If you think about it, I think you'll agree that any group that has
ordinary addition as the associated binary operation will have
identity element e = 0.

Can we do a similar thing for the integers under multiplication,
denoted (Z, *)?  How would we find an identity element?

We want to find an element e such that

e*x = x for all x in G.

This forces e = 1.

So for groups where the operation is normal multiplicitive of
numbers, e = 1.

Notice however, that (Z, *) does not form a group. It fails condition
three, since the inverse of the element 2 (i.e., 1/2) is not an
element of the group (it isn't an integer). In the integers, every
nonidentity element is not invertible.

The rational numbers, Q, almost form a group under multiplication.
The identity is still e = 1, and the inverse of p/q is q/p. The
problem occurs when p = 0.  In this case, 0/q = 0 is not invertible.

Thus (Q, *) isn't a group either. If we let Q* denoted Q - {0}, then
(Q*, *) is a group. The identity is 1 and the inverse of p/q is given
by q/p.

Another example of a group is GL(2, R) = the set of all 2x2 invertible
matrices whose coefficients come from the set of real numbers R = the
set of all 2x2 matrices with nonzero determinant whose coefficients
come from the set of real numbers R. The group operation here is
matrix multiplication which should be familiar from high school
algebra.

To find the identity element here, we are searching for a matrix I in
the group such that

I*X = X for all X in GL(2, R).

I is a 2x2 matrix so it has coefficients:

[ a  b ]
[ c  d ]

I is invertible so a*b - c*d is not zero.

similarly, X is a 2x2 matrix so it has coefficients:

[ w  x ]
[ y  z ]

X is invertible so w*z - x*y is not zero.

Performing the multiplcation gives:

[ a  b ]  [ w  x ]     [ w  x ]
[ c  d ]  [ y  z ] =   [ y  z ]

[ a*w + b*y  a*x + b*z ]    [ w  x ]
[ c*w + d*y  c*x + d*z ] =  [ y  z ]

But this is only true if all of the following are true:

a = 1
b*y = 0
b*z = 0
d = 1
c*w = 0
c*x = 0

The conditions b*y = 0 and b*z = 0 imply:

[ b = 0 or y = 0 ] and [ b = 0 or z = 0 ]

which forces:

b = 0 or [ y = 0 and z = 0 ]

If y = 0 and z = 0, then w*z - x*y is zero which cannot be. Thus it
must be the case that b = 0.

A similar argument shows that c = 0.

Thus

I =

[ 1 0 ]
[ 0 1 ]

This is the identity element for the group GL(2, R). It is the
familiar identity matrix from high school algebra. Perhaps now you
know why it was called the identity matrix. One final note - in
this group, A*B is not necessarily equal to B*A. So it is not
necessarily the case that the binary operation * must be commutative.
Sometimes it is and sometimes it isn't. When * is commutative, the
group G is often called an abelian group (named after the
mathematician Abel).

The reason courses that talk about groups are generally entitled
"Abstract Algebra" is that we are proving things abstractly. The
statement I helped you prove first works for *any* group G, regardless
of what the identity element in that group might be. This is what is
so powerful about abstract algebra. We are taking a generic object and
supposing it has certain properties and then we are proving things
about it in the most general case.

The strength of such a proof comes from the fact that once we find a
specific example of a group, we already know certain things about it
because we have proved that certain properties hold for *all* groups -
regardless of what binary operation we use and regardless of what
set we use.

That is why we just referred to the identity element in the proof we
did first. We didn't say e = 0 or e = 1 or e is the 2x2 identity
matrix. We just don't know which particular identity element is being
referred to because we don't know what the group is. But that's not
important. The idea is that by working abstractly (and intentionally
not saying what the set and binary operation are), we now know a very
special property that will hold in *every* group we encounter.

I hope this helps.  Please write back if you'd like to talk about
this some more.

- Doctor Paul, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Modern Algebra

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