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Best answer
40 votes
40 votes

Let $G$ with the set $\{1,3,5,7\}$ be a group with respect to $\otimes _{8}$ operation because all elements are less than $8$ and co-prime to $8.$

Let $G_1=\{1,3\}, G_2=\{1,5\}$

$G_1$ and $G_2$ are sub-groups of $G$ having identity element $1.$ We can make a composition table and cross verify.

  1. $G_1\cap G_2=\{1\}$ is a subgroup containing only identity element (Trivial Subgroup).
  2. $G_1\cup G_2=\{1,3,5\}$ 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 by
9 votes
9 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.
3 votes
3 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.

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