Question

If (h, •) and (k, •) are subgroups of (g, •), prove that (h n k, •) is a subgroup of (g, •). can the same be said for the union, hu k? prove or give a counterexample.

Answer 1

Let
`\eta\in H\cap K`. Since both
`H` and
`K` are subgroups of
`G`, we have


`\eta\in H\cap K\implies\begin{cases}\eta\in H\\\eta\in K\end{cases}\implies\eta\in G`

Because both
`H,K` are subgroups of
`G`, then
`(H\cap K,\bullet)` contains the identity element
`e`. Furthermore, there must be some element
`\eta^(-1)\in H` and
`\eta^(-1)\in K`, i.e.
`\eta^(-1)\in H\cap K`, such that
`\eta\bullet\eta^(-1)=e`.

Now let
`\eta_1,\eta_2,\eta_3\in H\cap K`. By the same reasoning as above it follows that each of these belong in
`G`, and since
`(G,\bullet)` is a group, we have
`\eta_1\bullet(\eta_2\bullet\eta_3)=(\eta_1\bullet\eta_2)\bullet\eta_3`, so
`H\cap K` is associative under
`\bullet`.

So
`H\cap K` contains the identity, is closed with respect to inverses, and is associative under
`\bullet`. Therefore
`H\cap K` must be a subgroup of
`G`.

In the case of union, this is not always the case. Consider the group
`(\mathbb Z,+)` with subgroups
`(H,+)` and
`(K,+)` where
`H` is the set of all integer multiples of 2 and
`K` is the set of all integer multiples of 3.

It's easy to show that
`H` and
`K` are indeed subgroups, but this is not the case for
`H\cup K`. We have


`H\cup K=\{0,\pm2,\pm4,\ldots\}\cup\{0,\pm3,\pm6,\ldots\}`

Take the elements
`2\in H` and
`3\in K`. Addition yields
`2+3=5`, but
`5\\ot\in H\cup K`, so
`H\cup K` is not closed under
`+`.