359 views
1 votes
1 votes

My question is about sub-groups.

I didn't understand how A non-empty subset H is a sub-group of a finite group G  if and only if for all a belongs to H, b belongs to H,  a*b belongs to H.

For example, (⎨1, 1/2 and 2⎬, *) is a group on multiplication but H = ⎨1, 1/2 ⎬ is not a sub-group. This is because there is no inverse for 1/2 in H.

However, the PDF linked to on the Gate CSE site under 'Set Theory and Algebra' says:

Could someone please tell me what this means/how this is true?

1 Answer

0 votes
0 votes

Your set = { 1, 2, 1/2 } 

on * operator 2 * 2 = 4 not in Set. Hence not closed => not Group => no question of finite subset.

Besides, Necessary condition for a subgroup is,

order(Subgroup) = order(Group) / k , k = {1,2,3,...}

Here order(G) = 3

Hence order(S) can be only 1 as 1 = 3 / 3

Example you chose is, Subgroup = { 1, 1/2 }, hence order(S) = 2

which does not divide 3. Hence not subgroup by LaGrange's theorem.

edited by

Related questions

0 votes
0 votes
1 answer
1
dileswar sahu asked Oct 11, 2016
306 views
In the above question my doubt is instead of under multiplication if we change under ADDITION then what is its value?plz someone explain details.
1 votes
1 votes
1 answer
2
Balaji Jegan asked Oct 26, 2018
279 views
1 votes
1 votes
1 answer
4