note. prove this question without langrange theorem prove that G cannot have a subgroup H with |G| > 2? note. prove this question without langrange theorem |H| =n-1 where n= Solution | G | = n > 2 Lets assume there exist a subgroup, H of G suchthat |H| = (n-1) Since, |G| >2, so H contains atleast one non-identity element. Since, |G| = |H| + 1 So, there exist one element a of G that is not in H. So, a 2 = e, Since if inverse of a is in H then since H is a sub-Group, so a must also belong to H. And This is not possible. And outside H, a is the only element of G. Let, h 1 be a non-identity element of H. If ah 1 H then there exist h 2 H suchthat ah 1 = h 2 =>   a = h 2 h 1 -1 H this is Not Possible. If ah 1 is not element of H then ah 1 = a    => h 1 = a -1 a = e, Not Possible. Hence such a subgroup is Not Possible. .

1. note. prove this question without langrange theorem
prove that G cannot have a subgroup H with |G| > 2? note. prove this question without langrange
theorem |H| =n-1 where n=
Solution
| G | = n > 2
Lets assume there exist a subgroup, H of G
suchthat |H| = (n-1)
Since, |G| >2, so H contains atleast one non-identity element.
Since, |G| = |H| + 1
So, there exist one element a of G that is not in H.
So, a 2
= e,
Since if inverse of a is in H then since H is a sub-Group, so a must also belong to H. And This is
not possible. And outside H, a is the only element of G.
Let, h 1 be a non-identity element of H.
If ah 1 H then there exist h 2 H
suchthat ah 1 = h 2
=>Â Â a = h 2 h 1
-1
H this is Not Possible.
If ah 1 is not element of H then
ah 1 = a   => h 1 = a -1
a = e, Not Possible.
Hence such a subgroup is Not Possible.