Answer to Question #118525 in Abstract Algebra for Jflows

Question

Which of the following listed below a subgroup of G, if H and K are two subgroups of a group G?
a.H\\(\\mathrm{\\cap}\\)K
b.H/K
c.H-K
d.H\\(\\cup\\)K

Answer 1

Correct option is (a).

Reason:

Let fix notation: $\\\leq$ denote subgroup.

Now, we have given $G$ is a group,$H\leq G \&K\leq G$ .

Claim:

$$H\cap K\leq G$$

Proof:

It is sufficient to show $H\cap K \leq G$ if we show $H\cap K$ closed under multiplication and taking inverse in it.

Let, for every $h,k\in H\cap K \implies h\in H \& h\in K$ also $k\in H \& k\in K$ ,thus

$$hk\in H\&hk\in K \hspace{1cm}(\because H,K\leq G)$$

Hence,

$$hk\in H\cap K$$

Which implies $H\cap K$ is closed under multiplication.

Now, consider

$$\forall h\in H\cap K\implies h\in H\&h\in K$$

Since, $H,K\leq G$ ,

$$h^{-1}\in H\&h^{-1}\in K\implies h^{-1}\in H\cap K$$

Thus, $H\cap K$ is closed under taking inverse.

Therefore,

$$H\cap K \leq G$$

For remaining 3 options we observe that if we choose $G=S_3$ and $H=\{e,(12)\}\&K=\{e,(13)\}$ , does not qualify the reaming options to be subgroup of $G$ .