Let G be a finite group of order n.Show for any field K ,G is isomorphic to a subgroup of GLn(K)?

Answer 1

(Remark: the fact that this is supposed to work with any field K is a hint that our matrices - elements of GLn(K) - can't be too complicated! In fact we can get away with using only 0s and 1s.)

The key is to think about the elements of G, call them g1, g2, ..., gn, also as permutations of the set { g1, g2, ..., gn } itself: any element gi permutes { g1, g2, ..., gn } to { g1 gi, g2 gi, ..., gn gi } (verify that that really is a permutation of the set!).

This is the key to showing that every finite group of order n is isomorphic to a subgroup of Sn, the symmetric group of order n!. But Sn itself is isomorphic to a subgroup of GLn(K), namely the subgroup of "permutation matrices", which have exactly one 1 in every row and column and 0s everywhere else.

Answer 2

once you write "in" interior the expression "H in ok in G", you probably mean that H is a subset of ok and ok is a subset of G. (the reason of being choosy is that "in" all via itself is commonly used to abbreviate elementhood, no longer set containment. The quite ambiguous word "is contained in" normally takes the two which ability. even though that's terrific to assert precisely what you mean.) i could first instruct that if A is a subgroup of a finite team B then |B:A| = |B|/|A|. To instruct this that's adequate to instruct that |B| = |B:A| |A|. that's commonly completed in proofs of Lagrange's theorem (which via itself is commonly basically the assertion that |A| divides |B|; yet maximum proofs of this set up the reality via exhibiting that |B| = |B:A| |A|). To summarize the argument, |B:A| is commonly via definition the form of left cosets of A in B (or the form of excellent cosets of A in B; it seems to be a similar selection). One then proves that each and each left coset of A in B has a similar cardinality as A (via exhibiting, e.g. that in the process the experience that your coset is bA, the map from A to bA given via sending x to bx is a bijection), and that the left cosets of A in B kind a partition of B (each and every b in B is in a coset of A in B, specifically the coset bA of A in B; and if bA and b'A are cosets of A in B that have nonempty intersection, a quick argument shows that they could desire to be a similar coset). each and every time a series S could be partitioned into N subsets of equivalent length ok you end that |S| = Nk, so which you practice that everyday actuality right here, with S = B, N = |B:A|, and ok = |A|. besides, enable's assume you have that result. The info then is going as follows. via fact ok is a subgroup of G you realize via employing the above to B = G and A = ok that |G:ok| = |G|/|ok|. via fact H is a subgroup of ok you realize via employing the above to B = ok and A = H that |ok:H| = |ok|/|H|. It follows from those 2 formula that |G:ok| |ok:H| = (|G|/|ok|) (|ok|/|H|) = |G|/|H|. and because H is a subgroup of G, the above with B = G and A = H tells us that |G|/|H| = |G:H|. consequently |G:H| = |G:ok| |ok:H|.

Answer 3

GLn(K) = nlog G ^K where K=1