If H and K are subgroups of group G, where e is the identity of G. Prove that
H nK is a subgroup of G.Also,if lHl = 10.and lKl = 21 Then HnK={e}.


### Proof: \( H \cap K \) is a Subgroup of \( G \)

To prove that \( H \cap K \) is a subgroup of \( G \), we use the **subgroup test**, which requires the following conditions:

1. **Identity Element**: \( e \in H \cap K \).
2. **Closure under Group Operation**: For all \( a, b \in H \cap K \), \( ab \in H \cap K \).
3. **Closure under Inverses**: For all \( a \in H \cap K \), \( a^{-1} \in H \cap K \).

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#### Step 1: \( e \in H \cap K \)
Since \( e \) is the identity element of \( G \), and both \( H \) and \( K \) are subgroups of \( G \), \( e \in H \) and \( e \in K \). Therefore, \( e \in H \cap K \).

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#### Step 2: Closure under Group Operation
Let \( a, b \in H \cap K \). This means \( a \in H \) and \( a \in K \), as well as \( b \in H \) and \( b \in K \). Since \( H \) and \( K \) are subgroups, \( ab \in H \) (closure in \( H \)) and \( ab \in K \) (closure in \( K \)). Therefore, \( ab \in H \cap K \).

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#### Step 3: Closure under Inverses
Let \( a \in H \cap K \). Then \( a \in H \) and \( a \in K \). Since \( H \) and \( K \) are subgroups, \( a^{-1} \in H \) (closure under inverses in \( H \)) and \( a^{-1} \in K \) (closure under inverses in \( K \)). Therefore, \( a^{-1} \in H \cap K \).

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Since \( H \cap K \) satisfies all three conditions, \( H \cap K \) is a subgroup of \( G \).

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### Second Part: If \( |H| = 10 \) and \( |K| = 21 \), then \( H \cap K = \{ e \} \)

We are given that the orders of the subgroups \( H \) and \( K \) are \( |H| = 10 \) and \( |K| = 21 \). The order of their intersection, \( |H \cap K| \), must divide the orders of both \( H \) and \( K \) (by Lagrange's Theorem).

#### Divisors of \( |H| = 10 \):
The divisors of 10 are: \( 1, 2, 5, 10 \).

#### Divisors of \( |K| = 21 \):
The divisors of 21 are: \( 1, 3, 7, 21 \).

#### Common Divisors:
The only common divisor of \( 10 \) and \( 21 \) is \( 1 \). 

Therefore, the order of \( H \cap K \) must be \( |H \cap K| = 1 \). Since the only subgroup of order 1 is the trivial subgroup \( \{ e \} \), we conclude that:

\[
H \cap K = \{ e \}.
\]

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### Conclusion

1. \( H \cap K \) is a subgroup of \( G \).
2. If \( |H| = 10 \) and \( |K| = 21 \), then \( H \cap K = \{ e \} \).

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Would you like further details or examples to clarify this proof? Here are five related questions:

1. How does Lagrange's theorem help in determining subgroup orders?
2. Can \( H \cap K \) ever have a non-trivial element if \( |H| \) and \( |K| \) are relatively prime?
3. What if \( |H| = 12 \) and \( |K| = 15 \)? What would \( |H \cap K| \) be?
4. How would the proof change if \( H \) and \( K \) were not subgroups but only subsets?
5. Can this result be extended to more than two subgroups (e.g., \( H \cap K \cap L \))?

**Tip**: Always remember that subgroup intersection orders must divide the individual orders of the subgroups, a key property derived from Lagrange's theorem.

If H and K are subgroups of group G, where e is the identity of G. Prove that H ∩ K is a subgroup of G. Also, if |H| = 10 and |K| = 21, then H ∩ K = {e}.

Learn how to prove that the intersection of subgroups H and K in a group G is a subgroup using the Subgroup Test. Additionally, explore why the intersection is trivial (H ∩ K = {e}) when |H| = 10 and |K| = 21, based on Lagrange's Theorem. Suitable for undergraduate mathematics students.

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