To prove that a subset $S$ of a group $G$ is a subgroup of $G$ if and only if:

1. $S$ is non-empty.
2. For any $x, y \in S$, $x^{-1}y \in S$.

We'll proceed with both directions: proving that these conditions are necessary and sufficient.

1. Necessary Condition

Assume $S$ is a subgroup of $G$. We need to prove that $S$ satisfies conditions (i) and (ii).

(i) $S$ is non-empty:

• Since $S$ is a subgroup, it must contain the identity element $e$ of $G$, because the identity element is in every subgroup by definition. Therefore, $S$ is non-empty.

(ii) For any $x, y \in S$, $x^{-1}y \in S$:

• $S$ being a subgroup means that $S$ is closed under the group operation and taking inverses.
• Since $x \in S$, we have $x^{-1} \in S$ because $S$ is closed under inverses.
• Since $S$ is closed under the group operation, $x^{-1} \cdot y \in S$.
• Therefore, condition (ii) is satisfied.

Thus, if $S$ is a subgroup of $G$, conditions (i) and (ii) hold.

2. Sufficient Condition

Now, assume that conditions (i) and (ii) hold. We need to show that $S$ is a subgroup of $G$.

(i) $S$ is non-empty:

• This condition is already given, so we know $S$ contains at least one element.

(ii) $x^{-1} y \in S$ for all $x, y \in S$:

• To show $S$ is a subgroup, we must prove that $S$ contains the identity element, is closed under the group operation, and is closed under inverses.

Identity element:

• Since $S$ is non-empty, take any element $a \in S$.
• Applying condition (ii) with $x = y = a$, we have $x^{-1} y = a^{-1} a = e \in S$.
• Therefore, $S$ contains the identity element $e$.

Closure under inverses:

• Take any $x \in S$. Using condition (ii) with $y = e$, we get $x^{-1} e = x^{-1} \in S$.
• So $S$ is closed under inverses.

Closure under the group operation:

• For $x, y \in S$, using condition (ii), we know $x^{-1} y \in S$.
• Let $z = x^{-1} y$. Since $x^{-1} y = z \in S$ and $x^{-1} \in S$, multiplying both sides by $x$ gives $y = x z$.
• Therefore, $S$ is closed under the group operation because $y = x \cdot (x^{-1}y) \in S$.

Since $S$ contains the identity, is closed under inverses, and is closed under the group operation, $S$ is a subgroup of $G$.

Conclusion

We have shown that $S$ is a subgroup of $G$ if and only if:

1. $S$ is non-empty.
2. For any $x, y \in S$, $x^{-1}y \in S$.

Would you like more details on any part of this proof?

Here are 5 related questions you might find interesting:

1. What is the difference between a normal subgroup and a subgroup?
2. How can we show that the intersection of two subgroups is a subgroup?
3. Can a single element in a group generate a subgroup?
4. What is the significance of the subgroup criterion in finite groups?
5. How does the subgroup criterion apply to the center of a group?

Tip: When verifying if a subset is a subgroup, checking closure under the group operation and inverses is often quicker if the identity is already known to be in the subset.