Let G be a group, and suppose that S is a subset of G. Prove that S is a subgroup of G if and only if S is non-empty, and for all x,y in S we have x^-1y in S

To prove that \( S \) is a subgroup of \( G \) if and only if \( S \) is non-empty, and for all \( x, y \in S \), we have \( x^{-1}y \in S \), we'll prove this in two parts: necessity and sufficiency.

### Part 1: Necessity (Subgroup \(\Rightarrow\) Condition)

Assume that \( S \) is a subgroup of \( G \). We need to show that \( S \) is non-empty and for all \( x, y \in S \), \( x^{-1}y \in S \).

1. **Non-empty:** Since \( S \) is a subgroup of \( G \), it must contain the identity element \( e \). Therefore, \( S \) is non-empty.

2. **For all \( x, y \in S \), \( x^{-1}y \in S \):** Since \( S \) is a subgroup, it must be closed under the group operation and inverses. This means that for any \( x \in S \), the inverse \( x^{-1} \in S \), and for any \( y \in S \), the product \( x^{-1}y \) must also be in \( S \). Therefore, \( S \) satisfies the given condition that for all \( x, y \in S \), \( x^{-1}y \in S \).

Hence, if \( S \) is a subgroup of \( G \), then \( S \) is non-empty and for all \( x, y \in S \), \( x^{-1}y \in S \).

### Part 2: Sufficiency (Condition \(\Rightarrow\) Subgroup)

Now, assume that \( S \) is non-empty and that for all \( x, y \in S \), \( x^{-1}y \in S \). We need to prove that \( S \) is a subgroup of \( G \). To do this, we need to verify the subgroup criteria:

1. **Identity element:** Since \( S \) is non-empty, take any element \( a \in S \). Consider the case where \( x = y = a \). Then, by the given condition, \( a^{-1}a = e \in S \). Therefore, \( S \) contains the identity element \( e \).

2. **Closure under inverses:** Let \( x \in S \). Since \( e \in S \), by the given condition, \( x^{-1}e = x^{-1} \in S \). Therefore, \( S \) is closed under taking inverses.

3. **Closure under the group operation:** Let \( x, y \in S \). We need to show that \( xy \in S \). Since \( y \in S \) and \( x^{-1} \in S \), by the given condition, \( (x^{-1})^{-1}y = xy \in S \). Hence, \( S \) is closed under the group operation.

Since \( S \) contains the identity element, is closed under taking inverses, and is closed under the group operation, \( S \) is a subgroup of \( G \).

### Conclusion:

Thus, \( S \) is a subgroup of \( G \) if and only if \( S \) is non-empty and for all \( x, y \in S \), \( x^{-1}y \in S \).

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Would you like further details or have any questions?

Here are 5 related questions for deeper exploration:

1. What are some examples of subgroups in well-known groups like \( \mathbb{Z} \) or \( \mathbb{R}^* \)?
2. How can you check if a given subset of a group is a subgroup using the two-step subgroup test?
3. How does the property \( x^{-1}y \in S \) relate to the concept of cosets in group theory?
4. Can the condition \( x^{-1}y \in S \) be interpreted geometrically for certain groups, such as matrix groups?
5. How does this subgroup criterion simplify in the case of finite groups?

**Tip:** When dealing with subgroups, remember that the identity element is always a part of any subgroup, and this is often a quick way to check for non-empty subsets.

Let G be a group, and suppose that S is a subset of G. Prove that S is a subgroup of G if and only if S is non-empty, and for all x,y in S we have x^-1y in S.

Explore the proof that a subset S of a group G is a subgroup if and only if S is non-empty and for all x, y in S, x^(-1)y is in S. This problem highlights key concepts in group theory, such as closure under the group operation and the inverse element. Suitable for university-level students studying Abstract Algebra.

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