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**Let \( H \) and \( K \) be two subgroups of a group \( G \). Then for any \( a \in G \), show that:**

\[
(H \cap K)a = Ha \cap Ka.
\]

### Proof:

We need to show that \( (H \cap K)a = Ha \cap Ka \), where \( H \cap K \) is the intersection of the subgroups \( H \) and \( K \), and \( Ha \), \( Ka \) are the left cosets of \( H \) and \( K \) by \( a \), respectively.

#### Step 1: Show \( (H \cap K)a \subseteq Ha \cap Ka \)

- Suppose \( x \in (H \cap K)a \). 
- This means that there exists some \( h \in H \cap K \) such that \( x = ha \). 
- Since \( h \in H \cap K \), by definition, \( h \in H \) and \( h \in K \).
- Therefore, \( x = ha \in Ha \) because \( h \in H \), and similarly \( x = ha \in Ka \) because \( h \in K \).
- Hence, \( x \in Ha \cap Ka \).

Thus, \( (H \cap K)a \subseteq Ha \cap Ka \).

#### Step 2: Show \( Ha \cap Ka \subseteq (H \cap K)a \)

- Suppose \( x \in Ha \cap Ka \).
- This means that there exist \( h_1 \in H \) and \( k_1 \in K \) such that \( x = h_1a = k_1a \).
- So, \( h_1a = k_1a \), which implies \( h_1 = k_1 \), since \( a \) is cancellable in group \( G \).
- Thus, \( h_1 = k_1 \in H \cap K \).
- Therefore, \( x = h_1a = (h_1 \in H \cap K)a \).
- Hence, \( x \in (H \cap K)a \).

Thus, \( Ha \cap Ka \subseteq (H \cap K)a \).

#### Conclusion:

Since we have shown both inclusions \( (H \cap K)a \subseteq Ha \cap Ka \) and \( Ha \cap Ka \subseteq (H \cap K)a \), we conclude that:

\[
(H \cap K)a = Ha \cap Ka.
\]

---

Would you like a further breakdown of the steps or any more details? Here are 5 related questions to deepen understanding:

1. What properties of subgroups make the proof above valid?
2. How does the left coset property impact the relationship between subgroups?
3. Can this proof be extended to right cosets? Why or why not?
4. What role does the cancellability of \( a \) in \( G \) play in this proof?
5. How would the proof change if \( H \) and \( K \) were not subgroups but normal subgroups?

**Tip:** When working with cosets and subgroup intersections, always check if each set is closed under the group operation for easier manipulation!

Let H and K be two subgroups of a group G. Then for any a ∈ G, show that (H ∩ K)a = Ha ∩ Ka.

This solution walks through proving that the intersection of subgroups H and K, when left-multiplied by an element a in group G, is equal to the intersection of the left cosets Ha and Ka. The proof explores key concepts of coset properties and subgroup intersections in group theory.

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