# Closure of groups

Let G be a group and my book defines closure as: For all a,bε G the element a*b is a well defined element of G. Then G is called a group. When they say well defined element does that mean I have to show a*b is well defined and it is a element of the group? Or do I just show a*b is closed under *(the operation)?

I've always found the closure axiom a bit silly. It's implied if you just write $*: G\times G\to G$. All it means is that, given $a,b\in G$, there's a thing named $a*b$, and that whatever this thing is, it belongs to $G$.

Thanks. I saw that other abstract algebra books have it defined as how you said it. My book apparently has it defined a little differently.

This will often depend on the context. * is by definition a binary function *:GxG→G, and a function is well-defined by definition.. I think the problem is best illustrated by examples:

Let G = {[x]: x is an integer not a multiple of 3}, where [x] = {integers y s.t. x~y}, where we write x~y if x and y leave the same remainder upon division by 3 (alternatively, x-y is divisible by 3). The elements of G are sets called equivalence classes of Z modulo 3, and we can easily verify that G={,}. Now define a binary operation on G by [a]*=[ab]. At first glance it might not be obvious that * is well-defined since [a] and have many different representations, and [ab] might depend on which of these representations we choose. For example = and =, so we better make sure that we get *=* with how we defined *! We can verify that *===*, since 155 leaves remainder 2 upon division by 3. Indeed, we can prove that [a]*=[ab] gives the same element of G no matter how we choose to write [a] and , i.e. * is well defined!

Edit: G={,,} is not a group ( is not invertible), edited so that G={,}

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Yes its just like showing a function is well defined. I wasn't sure if it just suffices to show that (G,*) is closed under the operation *.