I've got a question on abstract algebra?Pls help?10 PTS!

< (is a subgroup of),
[G:H] (the index of H
in G)
1. If H show that if G is
finite, then [G:H] =
[K:G] [G:K].
2. if P is a prime
number and a is an
integer, show that
a^p = a (mod P),
given G = finite
group and a £ G.
3. Prove that if H
and K are
subgroups of a
group, then H
intersection K is
also a subgroup of
G.

Pls, you guys really
need to help me on
this!

Correct me if I've gotten the question wrong. For 1, the way you've asked the question must have some sort of typo so I've attempted to correct it.

1: If H is normal in K and K is normal in G then if G is finite,
[G:H] = [G:K] [K:H]

Proof: Lagrange's theorem states the following:
[G:H] = |G|/|H|
[G:K] = |G|/|K|
[K:H] = |K|/|H|

it follows that:
[G:K] * [K:H] = |G|/|K| * |K|/|H| = |G|/|H| = [G:H]

2: if P is a prime number and a is an integer, then a^p = a (mod P)
Proof:
say that a = 0 (mod P). Then a to any power is the same.
say that a ≠ 0 (mod P)
multiplication mod P over {1,2,3,... p-1} forms a group (in modulo p, multiplication is invertible).
By Lagrange's theorem, the order of a (call it m) must divide the order of the group (p - 1), and a^m = 1 (mod p). Thus, we can say that p-1 = k*m for some integer k. It follows that
a^(p-1) = a^(km) = (a^m)^k = 1^k = 1 (mod p)
QED

3: In order to show that a given subset is a subgroup, you have to show that it has closure, an identity element, and inverses.

CLOSURE: Take a and b to be any two (not necessarily but possibly distinct) elements in the intersection of H and K. H is a subgroup, which means that a*b is in H. K is a subgroup, so a*b is in K. Since a*b is in both H and K, it is in their intersection. Thus, the subset is closed under the group operation

IDENTITY: H and K are both subgroups, which means they both contain the identity element. Thus, the identity is in the intersection.