# Basic Group Question

• Dec 14th 2011, 10:39 AM
wsldam
Basic Group Question
Let S be a non-empty set with a binary operation that satisfies the following:

1) the operation is associative
2) there exists a right sided identity
3) there exists a right sided inverse

Is S a group?

It seems like this should be simple but I can neither produce a counter nor adequately show it is true (the couple of 'proofs' I made seemed a bit hand-wavy)
• Dec 14th 2011, 11:25 AM
Deveno
Re: Basic Group Question

but why? why does "one-sided" (in this case, right) identity and inverse, imply "two-sided"?

first of all, it's important to realize that we need both a right-identity AND a right-inverse for this to work.

let's call our (right) identity, e'. so all we know, so far, is that for any a in S, a*e' = a.

but...we also have a right-inverse for a, that is, some b for which a*b = e'.

now, suppose for a moment that a*a = a (for some particular a). let b be a right inverse for a.

then (a*a)*b = a*b = e', so
a*(a*b) = e' (by associativity)
a*e' = e' (since a*b = e')
a = e'. remember this fact.

now, since every element has a right-inverse, so does e', that is, there is some f, for which:

e'*f = e'.

but then:

f = f*e' = f*(e'*f) = (f*e')*f = f*f, so f = e' (from our "remembered fact").

so e' is its own right inverse.

again, let's chose a and b so that a*b = e'.
now b*a = (b*e')*a (since e' is a right-identity)
= (b*(a*b))*a (since a*b = e')
= ((b*a)*a)*a = (b*a)*(b*a), by associativity (twice).

thus b*a = a*b = e'.

now, if b is such that b*a = a*b = e', then:

e'*a = (a*b)*a = a*(b*a) = a*e' = a, so e' is also a left-identity.

is this two-sided identity unique?

suppose we had another, called e".

then e' = e'*e" = e" (since both are two-sided identities).

so our two-sided identity is unique, and we can just call it "e".

is the element b for which b*a = a*b = e unique?

suppose that b' is another such element. so b'*a = a*b' = e, as well.

then b' = b'*e = b'*(a*b) = (b'*a)*b = e*b = b.

so inverses are unique as well.
• Dec 14th 2011, 11:28 AM
wsldam
Re: Basic Group Question
By the way here are the proofs I roughed out (e is the identity and z is the inverse):

For identity:
$a \star e = a$ (known)
$(a \star e) \star a = a \star a$
$a \star (e \star a) = a \star a$ (associative)
Because a=a, this implies $e \star a = a$ (possible hand waving)

For inverse:
$a \star z = e$ (known)
$(a \star z) \star a = e \star a$
$(a \star z) \star a = a \star e$ (by previous hand waving)
$a \star (z \star a) = a \star e$ (associative)
Because a=a, this implies $z \star a = e$ (more possible hand waving)
• Dec 14th 2011, 11:37 AM
Deveno
Re: Basic Group Question
Quote:

Originally Posted by wsldam
By the way here are the proofs I roughed out (e is the identity and z is the inverse):

For identity:
$a \star e = a$ (known)
$(a \star e) \star a = a \star a$
$a \star (e \star a) = a \star a$ (associative)
Because a=a, this implies $e \star a = a$ (possible hand waving)

For inverse:
$a \star z = e$ (known)
$(a \star z) \star a = e \star a$
$(a \star z) \star a = a \star e$ (by previous hand waving)
$a \star (z \star a) = a \star e$ (associative)
Because a=a, this implies $z \star a = e$ (more possible hand waving)

with the first one, you can't "cancel a on the left" until you know every a has a left-inverse. the same applies to the second "proof".
• Dec 14th 2011, 02:49 PM
Drexel28
Re: Basic Group Question
I would like to comment, that while this theorem may seem very stupid, in the sense that it isn't useful there is a canonical application of this theorem (most particularly the weakening to existence of only left inverses): a unital ring $D$ is a division ring if and only if the only left ideals of $D$ are $(0),(1)$ (same goes for right).