Question

Let \( G \) be a group with identity element \( e \), and let \( g, h \in G \) be such that the following hold: \[ g \neq e, \quad g^2 = e, \quad h \neq e, \quad h^2 \neq e, \quad {and} \quad ghg^{-1} = h^2. \] Then, the least positive integer \( n \) for which \( h^n = e \) is (in integer).

Hint:

For elements in groups where conjugation by one element doubles the powers of another element, track the powers and find the smallest \( n \) that brings the element back to the identity.

Answer 1

Step 1: Analyzing the Group Properties
We are given the following:

• \( g^2 = e \), so \( g \) is an element of order 2.
• \( ghg^{-1} = h^2 \), so conjugation by \( g \) maps \( h \mapsto h^2 \).

Step 2: Iterative Conjugation
We observe how \( h \) transforms under repeated conjugation by \( g \): 

\[ ghg^{-1} = h^2, \quad g^2 h g^{-2} = g (h^2) g^{-1} = (gh^2g^{-1}) = (ghg^{-1})^2 = (h^2)^2 = h^4, \] \[ g^3 h g^{-3} = g(h^4)g^{-1} = (gh^4g^{-1}) = (ghg^{-1})^4 = (h^2)^4 = h^8, \] and so on. 

Hence, by induction, we find: \[ g^n h g^{-n} = h^{2^n}. \] Step 3: Using Conjugation to Find the Order of \( h \)
Now consider the identity: \[ g^n h g^{-n} = h^{2^n}. \] So, if \( h^{2^n} = h \), then: \[ h^{2^n - 1} = e. \] Therefore, the order of \( h \) divides \( 2^n - 1 \). To find the least such positive integer \( n \) such that \( h^n = e \), we try successive powers. Let’s suppose \( h \neq e \), and find the smallest \( n \) such that \( h^n = e \), under the rule \( ghg^{-1} = h^2 \). Then: \[ ghg^{-1} = h^2 \Rightarrow g h^k g^{-1} = h^{2k}. \] That means conjugation doubles the exponent. Let’s suppose \( h^n = e \), and try to find the smallest such \( n \). Suppose \( h^3 = e \). Then \( h \) has order 3. Then: \[ ghg^{-1} = h^2, \quad g h^2 g^{-1} = (ghg^{-1})^2 = (h^2)^2 = h^4 = h \Rightarrow h^3 = e. \] This is consistent. Thus, \( h \) having order 3 is consistent with all given information. 
Step 4: Conclusion
The least positive integer \( n \) such that \( h^n = e \) is: 

\[ \boxed{3} \]