15. Let G be a group, and suppose that S is a subset of G. Prove that S is a subgroup of G iff (i) S is non empty and (ii) for x,y in S, we have x^-1y is in S

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.

15. Let G be a group, and suppose that S is a subset of G. Prove that S is a subgroup of G iff (i) S is non-empty and (ii) for x,y in S, we have x^-1y is in S.

Explore the proof that a subset S of a group G is a subgroup if and only if it is non-empty and for any x, y in S, x^-1y is in S. This problem delves into key concepts in group theory, including the subgroup criterion. Suitable for undergraduate students studying abstract algebra.

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