https://file.aiquickdraw.com/mathbot/file/1730933000414jknm15tz.jpg

To prove that the necessary and sufficient condition for a non-empty subset \( H \) of a group \( (G, \ast) \) to be a subgroup is:

\[
a, b \in H \Rightarrow a \ast b^{-1} \in H,
\]

we need to verify both directions of the implication: necessity and sufficiency.

### Sufficiency
Assume that \( H \) is a non-empty subset of \( G \) and satisfies the condition:
\[
a, b \in H \Rightarrow a \ast b^{-1} \in H.
\]
To show that \( H \) is a subgroup of \( G \), we need to demonstrate that \( H \) satisfies the subgroup criteria:
1. **Closure under the group operation:** If \( a, b \in H \), then \( a \ast b \in H \).
2. **Existence of the identity element in \( H \):** There exists an element \( e \in H \) such that for any \( a \in H \), \( e \ast a = a \ast e = a \).
3. **Existence of inverses for each element in \( H \):** For each \( a \in H \), there exists an \( a^{-1} \in H \) such that \( a \ast a^{-1} = e \).

Since \( H \) is non-empty, let \( a \in H \). Setting \( a = b \) in the condition \( a, b \in H \Rightarrow a \ast b^{-1} \in H \), we get:
\[
a \ast a^{-1} \in H.
\]
Since \( a \ast a^{-1} = e \), where \( e \) is the identity element of \( G \), we conclude that \( e \in H \).

Now, to show closure, let \( a, b \in H \). Then, using the given condition, we have:
\[
a \ast b^{-1} \in H.
\]
Since \( e \in H \), taking \( a = e \) in the condition implies that for any \( b \in H \), \( b^{-1} \in H \). Therefore, \( H \) contains the inverse of each of its elements.

Finally, for closure under the operation, consider any \( a, b \in H \). We have shown that \( b^{-1} \in H \), so by applying the condition again, \( a \ast b = a \ast (b^{-1})^{-1} \in H \).

Thus, \( H \) is a subgroup of \( G \) since it satisfies the subgroup criteria.

### Necessity
Now, we prove that if \( H \) is a subgroup of \( G \), then it satisfies the condition \( a, b \in H \Rightarrow a \ast b^{-1} \in H \).

If \( H \) is a subgroup, then by definition:
1. \( H \) contains the identity element \( e \).
2. \( H \) is closed under the group operation \( \ast \).
3. For any \( a \in H \), \( a^{-1} \in H \).

Now, if \( a, b \in H \), then \( b^{-1} \in H \) because \( H \) is closed under inverses. Since \( H \) is also closed under the group operation, \( a \ast b^{-1} \in H \).

Thus, the condition \( a, b \in H \Rightarrow a \ast b^{-1} \in H \) holds.

### Conclusion
We have shown that the condition \( a, b \in H \Rightarrow a \ast b^{-1} \in H \) is both necessary and sufficient for \( H \) to be a subgroup of \( G \).

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

Here are some related questions for further exploration:

1. How does this criterion compare to the three-step subgroup test for subgroups?
2. Can this criterion be generalized to non-abelian groups as well?
3. Why does this criterion require the set \( H \) to be non-empty?
4. What changes, if any, are needed for this proof if \( G \) is not finite?
5. How does this criterion relate to the definition of a normal subgroup?

**Tip:** Remember, testing if a subset \( H \) satisfies \( a, b \in H \Rightarrow a \ast b^{-1} \in H \) can sometimes simplify subgroup verification compared to checking each subgroup property individually.

Prove that the necessary and sufficient condition for a non-empty subset H of a group (G, *) to be a subgroup is a, b ∈ H => a * b⁻¹ ∈ H.

Explore the proof for the necessary and sufficient condition for a subset H of a group (G, *) to be a subgroup: if a, b ∈ H, then a * b⁻¹ ∈ H. This proof is essential for understanding subgroup properties and group theory, suitable for undergraduate students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation