Question

Let be a finite group of order at least two and let e denote the identity element of G. Let σ: G →g be a bijective group homomorphism that satisfies the following two conditions:
(i): if σ(g)=g for some gεG, then g=e, 
(ii) (σ o σ) (g)=g for all g σ G. 
The n which of the following is/are correct ?

• A. For each gεG, there exists hεG such that h-1σ(h)=g
• B. There exists xεG such that xσ(x)≠e.
• C. The map satisfies σ(x)^-1 for everyone xεG
• D. The order of the group G is an odd number

Answer 1

The Correct Option is A, C, D

To solve the problem, we need to analyze the given statements about the group \( G \) and the homomorphism \( \sigma \). Let's break down the problem step-by-step:

1. Understanding the homomorphism \( \sigma: G \to G \):
   • \( \sigma \) is bijective, which means that it is both injective and surjective, implying a perfect one-to-one correspondence between elements of \( G \).
   • \( \sigma(g) = g \) if and only if \( g = e \) (identity element of \( G \)). Hence, the only element that is mapped to itself is the identity.
   • \( (\sigma \circ \sigma)(g) = g \) for all \( g \in G \). This implies that \( \sigma \) is an involution, meaning applying it twice will return the original element.
2. Investigating the options:

Option 1:

For each \( g \in G \), there exists \( h \in G \) such that \( h^{-1}\sigma(h)=g \)

• Since \( \sigma \) is bijective, for any \( g \in G \), we can find some \( h \in G \) such that \( \sigma(h) = h \cdot g \). This rearranges to \( h^{-1}\sigma(h) = g \) by the property of group operations.
• This option is correct.

Option 2:

There exists \( x \in G \) such that \( x\sigma(x)\neq e \)

• By investigating the properties of homomorphisms and involutions, generally there may or may not exist such an \( x \) specifically. However, reflexivity cannot be directly denied without additional constraints, hence this doesn't hold for every structure of \( G \).
• This option is incorrect based on general involution property.

Option 3:

The map satisfies \( \sigma(x)^{-1} \) for everyone \( x \in G \)

• It's not clear what exactly is meant here, but understanding the context: If \( \sigma \) indeed reverses every element to its inverse while acting as an involution, this can be interpreted as \( \sigma(x) = x^{-1} \), which might not hold universally.
• This option seems incorrect or incomplete without further context.

Option 4:

The order of the group \( G \) is an odd number

• Since each element \( \sigma \) is bijective and squares to the identity, every non-identity element \( g \in G \) can be paired with its inverse which cannot equal itself unless it’s order 2 or the identity.
• If the order of every non-identity element is not 2, the group must have an odd number of elements because they can be paired uniquely. This implies that \( |G| \) is odd.
• This option is correct.

In conclusion, the correct statements are:

• For each \( g \in G \), there exists \( h \in G \) such that \( h^{-1}\sigma(h)=g \).
• The order of the group \( G \) is an odd number.