Thread: groups, subgroups and the finite hypothesis

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?

For example take the group and take the infinite subset , is a subgroup?

Does doing induction over the size of the subgroup work?

(N, +) isn't be a subgroup because it isn't a group. (No inverses and no identity.)

How do you make the right Z and N for integers and naturals?

Originally Posted by amyw

(N, +) isn't be a subgroup because it isn't a group. (No inverses and no identity.)
How do you make the right Z and N for integers and naturals?

[TEX]\mathbb{Z}~\&~\mathbb{N}[/TEX] gives .
On the tool bar the gives [TEX][/TEX] wrap.

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.

Let <G, *> be a group, x, a, b ∈ G and e is the identify element. Prove that if a2 = e and a*x*a = x3, then x8 = e.