For G a group, it can be shown that a finite subset S ⊆ G is a subgroup if S is closed under multiplication. Use an example of an infinite group to show the above is not a sufficient "test" of closure. That is, show there is at least one counterexample to the above. But I'm a bit confused, S is a subgroup if it's closed under the operation of G, which implies G = (A,*) for some set A. So I can only use an example which is a multiplicative group, and infinite?
Duh! Let G = (R, *) where R is the set of nonzero real numbers. Then a subset, H ={ x ∈ G | x = 1 or x is irrational } is not a subgroup under multiplication since √3 ∈ H, but √3*√3 = 3 ∉ H. Hence multiplication in H is not closed. And that's the counterexample . . .?
No, multiplication means the group operation, whatever it is. For example, consider the group \((\mathbb{Z}, +)\) (integers under addition). \(\mathbb{N} = \lbrace 1, 2, 3, \ldots \rbrace\) is a subset of \(\mathbb{Z}\) that is closed under the group operation of addition. However, it is obviously not a subgroup (the identity element 0, and the inverses of the positive integers are missing)
A subgroup of (Z, +) would be (nZ, +) where n is any integer. 0Z would be a trivial group of 1 element. 2Z would be the subgroup of even numbers closed under addition.
Correct. Whereas if we take the set of odd integers, this is still closed under addition, but is not a subgroup (it is, in fact, a coset of the subgroup \((2\mathbb{Z}, +)\)). Thus, closure under the group operation is not enough to ensure that a subset is a subgroup (except when the subset is finite).
Whoops, right you are! In fact, any coset of a subgroup \(H\) other than \(H\) itself is never closed under the group operation. Short proof: Let \(gH\) be a coset of \(H\) that is closed under the group operation, and let \(gh_1, gh_2 \in gH\) (where \(h_1, h_2 \in H\)). Closure implies: \(gh_1gh_2 \in gH \Rightarrow \\ gh_1gh_2 = gh_3,\ \exists h_3 \in H \Rightarrow \\ g = h_1^{-1}h_3h_2^{-1} \Rightarrow \\ g \in H \Rightarrow \\ gH = H\) Thanks for catching that, rpenner!
Well, I considered that "mulitiplication" in the question might mean the group operation, but I believed, incorrectly, that it was asking about multiplicative groups. I'll be careful not to repeat that mistake in the test, or the exam.
Referring to a group as "multiplicative" affects only the notation used when talking about it, not its properties. We call a group multiplicative only to make it understood that the group operation will be denoted by a dot (\(\cdot\)), or will be omitted altogether (as is common with our "usual" multiplication of real numbers, etc.). One other possible interpretation is in the context of rings or fields. If a ring or a field is already defined, we may say "multiplicative group" to mean the group formed by the set of all non-zero elements of the ring or field together with the multiplication defined on it.
Ok, so G is a set plus a binary operation, usually called multiplication. So if the operation is ordinary addition that's made clear by the context. But the binary operation might include both addition and multiplication (in the arithmetic sense), for instance if the set is matrices with entries from Z[sub]n[/sub], you use addition and multiplication (mod n). Or if the operation is defined as, say, x o y = x + y - xy, etc. The example of (Z, +) which has subsets {0 + 2Z} and {1 + 2Z} where the first is closed but the second isn't is probably what I should have had in my answer. Plus, I think the question could have been phrased more exactly, instead of "closed under multiplication" it should have said: "closed under the group operation". Except, of course, I should have known what was meant.