GATE CSE
First time here? Checkout the FAQ!
x
+6 votes
331 views

Let $G_1$ and $G_2$ be subgroups of a group $G$.

  1. Show that $G_1 \cap G_2$ is also a subgroup of $G$.
  2. Is $G_1 \cup G_2$ always a subgroup of $G$?.

 

asked in Set Theory & Algebra by Veteran (66.1k points) 1148 2196 2522 | 331 views

4 Answers

+5 votes
Best answer

G={1,3,5,7} is a group wrt $\bigotimes _{8}$ Opeartion bcz all elements are less than 8 and co-prime to 8.

Let G1={1,3} G2={1,5}

each are sub-group of G having identity element 1. we can make composition table and cross verify.

a)G1$\cap$G2={1} is a subgroup containing only identity element.(Trivial Subgroup).

b)G1$\cup$G2={1,3,5} which is not a subgroup.

Reason when we draw composition table 3*5=15 mod 8=7 which is not present in this subgroup so not satisfying closure property. Hence not a subgroup.

answered by Veteran (20.3k points) 13 78 206
edited by
Subgroups are closed under intersection but not under union.

@Rajesh Pradhan, Good counter example for Union, but for proof of 1st part, One may refer to link below.
http://math.stackexchange.com/questions/297356/intersection-of-2-subgroups-must-be-a-subgroup-
proof
+4 votes
ans for B:

No, G1 union G2 is not subgroup of G always.

G1 union G2 is subgroup of G if and only if either G1 is contained in G2 or G2 is contained in G1.
answered by Boss (8.3k points) 7 77 136
Didn't get, lets say G={a,b,c,d}

G1={b,c}

G2={c,d}

G1 U G2={b,c,d} is a subgroup but neither G1 is contained in G2 nor G2 is contained in G1.

Also it has not asked for proper subgroup, even G1 U G2 = G then also Gi U G2 is a subgroup.

Hence it should be always a subgroup.
R u applied group properties?
ok,I got it, such a silly mistake..thankuu
+2 votes

a) Let a,b $\epsilon$ (G1 $\cap$ G2)

Now, a,b $\epsilon$ G1 $\Rightarrow$ a*b $\epsilon$ G($\because$ G1 is algebraic structure)

Similarlly a-1, b-1 $\epsilon$ G1 ($\because$ G1 is group)

Also this will be true for G2.

$\therefore$ As a,b $\epsilon$ (G1 $\cap$ G2), a-1, b-1 $\epsilon$ (G1 $\cap$ G2), which will make it subgroup of G.

 

b) G$\cup$ Gneed not be a subgroup.

For example, Let G = {1,3,5,7} is a group w.r.t multiplication modulo 8.

Consider G1 = {1,3} and G2 = {1,5}

Here G& Gare subgroups of G but G$\cup$ G2 is not a subgroup of G.

answered by Loyal (4.4k points) 5 17 33
0 votes

 

Proof for part (a)

answered by Boss (8.8k points) 3 8 12


Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
Top Users Oct 2017
  1. Arjun

    23210 Points

  2. Bikram

    17018 Points

  3. Habibkhan

    6652 Points

  4. srestha

    5864 Points

  5. Debashish Deka

    5430 Points

  6. jothee

    4908 Points

  7. Sachin Mittal 1

    4762 Points

  8. joshi_nitish

    4274 Points

  9. sushmita

    3954 Points

  10. Silpa

    3698 Points


Recent Badges

Regular Juhi Sehgal
Popular Question vineet.ildm
Nice Comment Arjun
100 Club vipul verma
Notable Question jothee
Popular Question jothee
Nice Question shivangi5
Regular rinks5
Notable Question shipra tressa
Regular sasi
27,247 questions
35,056 answers
83,703 comments
33,183 users