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.