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**3.** Let \( a \) and \( b \) be two elements of a group. If the order of \( a \), denoted by \( o(a) = d \), then show that \( o(a^{-1}) = o(bab^{-1}) \).

### Solution:

We are asked to prove two things:
1. The order of \( a^{-1} \) is the same as the order of \( a \), i.e., \( o(a^{-1}) = o(a) \).
2. The order of \( bab^{-1} \) is also the same as the order of \( a \), i.e., \( o(bab^{-1}) = o(a) \).

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#### Part 1: Show \( o(a^{-1}) = o(a) \)
Let \( o(a) = d \), meaning \( a^d = e \) (where \( e \) is the identity element of the group).

Now, we need to determine the order of \( a^{-1} \). The order of an element is the smallest positive integer \( k \) such that \( (a^{-1})^k = e \).

Using the properties of group elements, we know:
\[
a^{-1} \cdot a^{-1} \cdot \ldots \cdot a^{-1} \ (\text{d times}) = (a^{-1})^d = (a^d)^{-1} = e^{-1} = e
\]
Thus, \( (a^{-1})^d = e \), showing that the order of \( a^{-1} \) is also \( d \). Hence,
\[
o(a^{-1}) = o(a) = d
\]

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#### Part 2: Show \( o(bab^{-1}) = o(a) \)
Let \( c = bab^{-1} \), and we aim to show \( o(c) = o(a) \).

The order of \( c \) is the smallest integer \( k \) such that \( c^k = e \).

We compute \( c^k \):
\[
c^k = (bab^{-1})^k = ba^k b^{-1}
\]
For the order of \( c \), we need \( c^d = e \). Now,
\[
c^d = ba^d b^{-1} = beb^{-1} = bb^{-1} = e
\]
Thus, \( (bab^{-1})^d = e \), which shows that the order of \( bab^{-1} \) is also \( d \).

Hence, \( o(bab^{-1}) = o(a) = d \).

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### Conclusion:
We have shown that:
\[
o(a^{-1}) = o(bab^{-1}) = o(a) = d
\]

Would you like further details or have any questions about this solution?

Here are 5 related questions for practice:
1. What is the order of the identity element in any group?
2. If \( a^2 = e \), what can you say about the order of \( a \)?
3. Can two elements in a group have different orders?
4. Show that \( o(ba) = o(ab) \) for any two elements \( a \) and \( b \) in an abelian group.
5. Is the conjugate \( bab^{-1} \) always the same as \( a \) in any group?

**Tip:** The concept of conjugates plays a key role in understanding how elements interact in group theory.

Let a and b be two elements of a group. If o(a) = d, then show that o(a^{-1}) = o(bab^{-1}).

In this abstract algebra problem, learn how to prove that the order of an element and its conjugate are the same in a group. Suitable for university-level group theory courses.

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