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There's a set consisting of 2 elements: G = {a,b}. In this set we define an operation * in the following way:



The question says: "Show that (G, *) is a commutative group".

It's easy show that, if it really is a group, then it is an abelian one since:

$(a*b)*a=b*a=b$ and $a*(b*a)=a*b=b$, so $(a*b)*a=a*(b*a)$.

But to prove that it actually is a group I need to satisfy the rest 3 requirements, namely it needs to be associative and there need to be an identity element as well as an inverse element. I have a problem with the latter two.

As there are only 2 elements in this set, then one of them must be an identity and the other one an inverse.

Let's assume that a is an identity:

$$a*a=a$$ and $$b*a=b$$ - so it should be fine.

Then b must be an inverse, however that can't be true, given that:

$$a*b=b$$ (not the identity element a)

Meanwhile, b can't be an identity element because: $a*b=b\neq a$

So, it this group really not abelian despite what the way the problem was formulated suggests or am I missing something?

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No such thing as a single inverse element; rather an inverse for each element in the group. – tabstop Feb 2 '14 at 17:27
up vote 1 down vote accepted

You are missing something. Since $a*a=a$ and $a*b=b*a=b,$ then the identity element has to be $a.$ Since $b*b=a,$ then $b$ has to be its own inverse. (A group's identity element is always its own inverse. Why?)

What you need to do is show that $*$ is an associative operation on $\{a,b\}.$ The rest should follow quite readily.

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You're right. I missed the fact that an element can be an inverse of itself and that each of the elements has its own inverse. Thank you for the reminder and sorry for my English. – user3221028 Feb 2 '14 at 17:37

If $a$ is the identity, then you want to find elements $x$ and $y$ such that $b*x = a$ and $a*y=a$; then $b^{-1} = x$ and $a^{-1} = y$. There are not a lot of choices :)

You may be misunderstanding the definition of "inverse element". Each element must have an inverse; the inverse may be different for each element. So your statement above that $b$ cannot be an inverse does not really make sense.

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So does it mean that a is both an identity element of this group and its own inverse? Similarly b is the inverse of itself? – user3221028 Feb 2 '14 at 17:34
Yes, that's exactly right. – rogerl Feb 2 '14 at 17:37
For example, in the integers, zero is both the identity element and its own inverse. (Of course, that's not the group you're dealing with here). – rogerl Feb 2 '14 at 17:37

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