While browsing Group Theory I ( Suzuki ) I noticed the following proposition.

If G is a finite group and S is a subset of G then closure in S suffices for S to be a subgroup.

Proof:

S is a subgroup if for all a,b,c in S

( i ) ab in S - closure

( ii ) a(bc)= (ab)c - associativity

( iii ) e in S - has identity

( iv ) a^(-1) in S - has inverse

let's prove them one by one:

( i) is proposed to be true ;

( ii ) is true for G and thus true for S ;

( iii ) since G is finite there is an integer n such that a^n = e thus e in S

( iv) since a a^(n-1) = e all a have an inverse.

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When in exercises the word 'finite' is added to group like 'G is a finite group ' we know that for G the group axioms are true like (i) to (iv) above AND that there is an integer n such that for all g in G g^n = e.

2-2018 Teaching by misleading

2 months ago

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