For example take the group and take the infinite subset , is a subgroup?
Supposes (G, *) is a group, and H is a non-empty finite subset of G that is closed under *. Prove that H is a subgroup of G. Give an example to show that this is not true without the finite hypothesis.
To prove that H is a subgroup I need to show that H has an inverse and an identity. Since * is associative in G, then * is associative in H, true? How can I show H has an inverse and an identity? Also, how can I find an example that makes this not true?
ok, H is non-empty, right? so we can pick an element a in H.
let's define a function f from H to H by:
f(h) = a*h (this is in H because H is closed under *. it could be that a is the only element of H, in which case, we would have f(a) = a*a = a. it is a good exercise to convince yourself that in this case, a must be the identity of G).
what can we say about this function f?
well, suppose f(h) = f(h').
this means a*h = a*h'.
now we can view both of these as elements of G (since H is a subset of G), and IN G:
a^{-1}*(a*h) = a^{-1}*(a*h')
(a^{-1}*a)*h = (a^{-1}*a)*h' (by the associativity of * in G)
e*h = e*h'
h = h' (in G).
but if h = h' in G, certainly h = h' in H (h is always h, no matter if we consider it as in H, or the larger set G).
so f(h) = f(h') implies h = h', that is f is INJECTIVE (one-to-one).
since H is FINITE, f is BIJECTIVE, as well. in particular, f is thus surjective (onto).
so for any x in H, there is some (possibly different) h in H, with a*h = x.
but...a is certainly in H (remember we choose a as "some element" (which we KNOW exists) of H).
so there is some y in H with: a*y = a. hold that thought.
now we can also (using the same a) make ANOTHER function g:H-->H by setting g(h) = h*a.
i leave it to you that prove that this function g is ALSO bijective (the proof is "almost" the same).
so we can write any x in H as x = h*a, for some h in H.
now, a little "magic".
remember the y we found with the property that a*y = a? we're going to prove that x*y = x, for ANY x in H.
x*y = (h*a)*y (using the fact that g is surjective, so for any x in H, we have x = g(h) = h*a for some h in H)
= h*(a*y) (by associativity)
= h*a (because a*y = a, by our choice of y)
= x (ta da!)
so the element y is a "right-identity" for H.
ok, again, since f is surjective, there is some b in H, with f(b) = y (we already know that y is in H, right?).
so there is an element b in H, with a*b = y.
now, remember, a was chosen "arbitrarily", we can make similar functions to f and g for every single element of H.
so for every a in H, we can produce a "right-inverse" for a.
the next step, is to show that y is actually a "two-sided identity" and "b" (the right-inverse of a) is a "two-sided inverse for a".
using g this time, we know there is SOME z in H, with g(z) = a, that is: z*a = a.
and we also know (using f) that for any x in H, there is some h in H with f(h) = x, that is: a*h = x.
so for ANY x in H:
z*x = z*(a*h) = (z*a)*h = a*h = x.
so z is a "left-identity" for H. so:
z = z*y (since y is a right-identity)
= y (since z is a left-identity). thus the "left-identity" and "right-identity" of H are actually the same (we can just use y, we don't need z anymore).
in a similar way, we can show (and i urge you to do this on your own), that any element a in H has a "left-inverse" c, with c*a = y.
therefore:
c = c*y = c*(a*b) = (c*a)*b = y*b = b.
so the "left-inverse" of a, and the "right-inverse" of a are one and the same, which we can just call a^{-1}.
we are almost done: we know that y is an identity for H, and H has inverses with respect to this identity. however, we would like to be sure that, in fact:
y = e, the identity of G.
note that y is an element of G, with:
y*y = y (since y is an identity for H).
since y is an element of G, y has a "G-inverse" y^{-1} (we don't know YET that this is the same as the "H-inverse").
so (in G):
y^{-1}*(y*y) = y^{-1}*y
(y^{-1}*y)*y = e
e*y = e
y = e
this means that y is indeed the identity of G, and thus "H-inverses" are the same as "G-inverses". that is: (H,*) is a subgroup of (G,*) (using the same operation, *).
*****************
so, what might go wrong in the infinite case?
well, it's likely the only infinite group you can name off the top of your head is (Z,+), the integers under addition. let's pick some a in Z, and see what goes wrong with what we did above.
since Z is a "concrete group" (that we know and love so VERY much), we can pick "some actual element for a". let's pick a = 2.
so we have the function:
f:Z-->Z given by: f(n) = 2n, for n in Z.
as before, it is easy to see this function is injective (one-to-one). but...it's not surjective, we only get EVEN integers in the image of f. for example, there is NO integer n with:
2n = 3.
so, for a counter-example, let's pick G = Z. now at first, one might be tempted to choose H = 2Z (the subset of all even integers), but that doesn't work, because it turns out that H is indeed a subgroup of Z.
how about H = {n in Z: n > 0}?
clearly this is closed under addition: the sum of two positive integers is again a positive integer. but....no positive integer (element of H) has an additive inverse that is positive, and H doesn't even have an identity.