Pls, I really need help on this:
1. If and H and K are subgroups of G, show that if G is finite, then [G:H] = [K:H] [G:K].
2.If P is a prime number and a is an integer, show that a^p = a (mod P), given G is a finite group and a £ G.
Just those two. Thanks a lot!
1. If and H and K are subgroups of G, show that if G is finite, then [G:H] = [K:H] [G:K].
2.If P is a prime number and a is an integer, show that a^p = a (mod P), given G is a finite group and a £ G.
Just those two. Thanks a lot!
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1) Note that since G is finite, so are any subgroups of G.
Hence, [G:H] = |G|/|H|, [K:H] = |K|/|H|, and [G:K] = |G|/|K| by Lagrange.
==> [G:H] = |G|/|H| = (|K|/|H|) * (|G|/|K|) = [K:H] * [G:K].
2) If you are working mod p, then G can't be just any finite group G.
Assuming that G = Z_p, then this is straightforward.
If a = 0 mod p, then this is trivially true.
Otherwise, Z_p with multiplication is a cyclic group of order p-1.
So, Lagrange's Theorem yields a^(p-1) = 1 (mod p).
==> a^p = a (mod p).
I hope this helps!
Hence, [G:H] = |G|/|H|, [K:H] = |K|/|H|, and [G:K] = |G|/|K| by Lagrange.
==> [G:H] = |G|/|H| = (|K|/|H|) * (|G|/|K|) = [K:H] * [G:K].
2) If you are working mod p, then G can't be just any finite group G.
Assuming that G = Z_p, then this is straightforward.
If a = 0 mod p, then this is trivially true.
Otherwise, Z_p with multiplication is a cyclic group of order p-1.
So, Lagrange's Theorem yields a^(p-1) = 1 (mod p).
==> a^p = a (mod p).
I hope this helps!