MHB Proving Lagrange Theorem for Finite Group G

  • Thread starter Thread starter onie mti
  • Start date Start date
  • Tags Tags
    Lagrange Theorem
onie mti
Messages
42
Reaction score
0
given that G is a finite group.

1) if H is a subgroup of G then |H| divides |G|
2) if a in G the ord(a)/|G|

i could prove no 2 using no 1 where i said ord(a)=|<a>| and <a> is a subgroub of G so by 1
ord(a)/|G|how cAN I prove 1
 
Physics news on Phys.org
Re: langrange theorem

onie mti said:
given that G is a finite group.

1) if H is a subgroup of G then |H| divides |G|
2) if a in G the ord(a)/|G|

i could prove no 2 using no 1 where i said ord(a)=|<a>| and <a> is a subgroub of G so by 1
ord(a)/|G|how cAN I prove 1
Very briefly, the (left) cosets of H partition G into a number of sets all having the same size as H. So the order of G is the order of H times the number of cosets.
 
This is what you need to do to prove (1).

Step one:

Show that the map $f:H \to Ha$ for any $a \in G$, given by $f(h) = ha$ is bijective.

Step two:

Conclude that $|H| = |Ha|$ for all $a \in G$.

Step three:

Show that if $x \in Ha \cap Hb$, then $Ha = Hb$.

Step four:

Conclude that the distinct cosets of $H$ form a partition of $G$.

Step five:

Conclude that $G = H \cup Ha_1 \cup \cdots \cup Ha_{k-1}$ for $k$ distinct cosets of $H$ in $G$

(note that this uses the fact that $|G|$ is FINITE).

Step six:

Use steps four and five to conclude that:

$|G| = k\ast|H|$.
 

Similar threads

Back
Top