Question

(a) Let H and K be subgroups of a group G. Then show by an example that H$\cup$K need not be a subgroup of G.

Short answer

If H and K are subgroups of a group G, then their$\mathrm{H}\cup \mathrm{K}$ union is not a subgroup of G.

Step-by-step solution

To find two subgroups of a group

We have thedefinition of a subgroupthat,

Let (G,*) be a group under binary operation, say * A non-empty subset H of G is said to be a subgroup of G, if (H,*) is itself a group.

Consider a group G=(Z,+) under usual addition '+', then the subgroups of a group G are given by,

$$\begin{gathered} \mathrm{H}=2\mathrm{Z}=\{0,\pm 2,\pm 4,\pm 6,...\}\mathrm{and} \\ \mathrm{K}=3\mathrm{Z}=\{0,\pm 3,\pm 6,\pm 9,...\} \end{gathered}$$

To show H∪K is not a subgroup of G.

Now, we take the union of two subgroups H and K of G.

$\mathrm{H}\cup \mathrm{K}=2\mathrm{Z}\cup 3\mathrm{Z}=\{0,\pm 2,\pm 3,\pm 4,\pm 6,...\}$

Here, $2,3\in 2Z\cup 3Z$but $2+3=5\notin 2Z\cup 3Z$

Hence, we prove thatlocalid="1659267378694" $2Z\cup 3Z$is not a subgroup of G, that is, localid="1659267383037" $\mathrm{H}\cup \mathrm{K}$is not a subgroup of G.