529 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 $G$?.

0

This might help to solve ...

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.

edited
+8
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
0
nice and simple
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.
0
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.
0
R u applied group properties?
0
ok,I got it, such a silly mistake..thankuu

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.

+1 vote

Proof for part (a)