Let G be group, let H be a subgroup of G, and let a and b be elements of G. Prove that
|aH| = |Hb|.

Let \( G \) be a group, \( H \) be a subgroup of \( G \), and let \( a, b \in G \). We want to prove that the left coset \( aH \) and the right coset \( Hb \) of \( H \) in \( G \) have the same cardinality, i.e., \( |aH| = |Hb| \).

### Step-by-Step Proof:

#### 1. **Define the cosets:**
   - The **left coset** of \( H \) with respect to \( a \) is given by:
     \[
     aH = \{ ah \mid h \in H \}.
     \]
   - The **right coset** of \( H \) with respect to \( b \) is given by:
     \[
     Hb = \{ hb \mid h \in H \}.
     \]

We need to show that these two sets, \( aH \) and \( Hb \), have the same size (i.e., the same number of distinct elements).

#### 2. **Construct a bijection between \( aH \) and \( Hb \):**
   Consider the map \( f : aH \to Hb \) defined by:
   \[
   f(ah) = hb \quad \text{for all } h \in H.
   \]
   This map takes each element of the left coset \( aH \), which is of the form \( ah \), and maps it to an element of the right coset \( Hb \), namely \( hb \).

#### 3. **Show that \( f \) is well-defined:**
   For \( f(ah) = hb \) to be well-defined, we must check that if two elements in \( aH \) are equal, their images under \( f \) must also be equal.
   
   Suppose \( ah_1 = ah_2 \) for some \( h_1, h_2 \in H \). Then, by left-multiplying both sides by \( a^{-1} \), we get:
   \[
   h_1 = h_2.
   \]
   Therefore, \( f(ah_1) = f(ah_2) \), meaning the map is well-defined.

#### 4. **Show that \( f \) is injective (one-to-one):**
   Suppose \( f(ah_1) = f(ah_2) \), meaning:
   \[
   hb_1 = hb_2 \quad \text{for some } h_1, h_2 \in H.
   \]
   This implies that:
   \[
   h_1 = h_2.
   \]
   Therefore, \( f \) is injective.

#### 5. **Show that \( f \) is surjective (onto):**
   For any element \( x = hb \in Hb \), there exists a corresponding element \( x = ah \in aH \) under the map \( f \). Therefore, the map is surjective.

Since \( f \) is both injective and surjective, it is a bijection between the sets \( aH \) and \( Hb \). Therefore, the two sets must have the same cardinality.

#### Conclusion:
Since \( f \) is a bijection, we conclude that:
\[
|aH| = |Hb|.
\]

---

### Additional Exploration:

Would you like further details on any specific part of the proof? Here are some related questions that might help deepen your understanding of this topic:

1. How do cosets partition a group into disjoint subsets?
2. Can we apply a similar argument to prove that left cosets and right cosets of a normal subgroup have the same size?
3. How does the index of a subgroup relate to the number of cosets?
4. How can we prove that the left and right cosets of any subgroup \( H \) in \( G \) have the same cardinality using Lagrange’s Theorem?
5. What happens if \( H \) is a normal subgroup in terms of coset equivalence?

**Tip:** A good strategy for proving set equalities is to construct a well-defined bijection, which is often the simplest way to show two sets have the same cardinality.

Let G be group, let H be a subgroup of G, and let a and b be elements of G. Prove that |aH| = |Hb|.

Explore the proof that the cardinalities of left and right cosets, |aH| and |Hb|, are equal for any group G and subgroup H. This solution includes detailed steps in constructing a bijection between cosets and applies fundamental group theory concepts.

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