Question

How many group homomorphisms are there from \(\Z_2\) to S₅ ?

• A. 40
• B. 41
• C. 26
• D. 25

Answer 1

The Correct Option is C

To determine how many group homomorphisms exist from \( \mathbb{Z}_2 \) to \( S_5 \), we need to understand the structure of these groups and the properties of homomorphisms.

\( \mathbb{Z}_2 \) is the additive group of integers modulo 2. It has two elements: 0 and 1, where 0 is the identity element.

\( S_5 \) is the symmetric group on 5 elements, which includes all permutations of a 5-element set. The group \( S_5 \) has 120 elements.

A group homomorphism \( f: \mathbb{Z}_2 \rightarrow S_5 \) will have the following properties:

• Maps the identity element of \( \mathbb{Z}_2 \) (which is 0) to the identity element of \( S_5 \) (the identity permutation).
• The image of 1 under the homomorphism, \( f(1) \), can be any element in \( S_5 \) such that when it is composed with itself (since 1 + 1 = 0 in \( \mathbb{Z}_2 \)), it yields the identity permutation. These elements are involutions in \( S_5 \) (elements of order 1 or 2).

Now, let's count the number of such involutions in \( S_5 \):

• The identity permutation is an involution.
• A transposition, such as (1 2), is an involution. There are \( \binom{5}{2} = 10 \) transpositions.
• A product of two disjoint transpositions, such as (1 2)(3 4), is an involution. There are \(\frac{1}{2} \times \binom{5}{2} \times \binom{3}{2} = 15\) such permutations.
• Other permutations have order greater than 2 and are not involutions.

Therefore, the number of involutions in \( S_5 \) is \( 1 + 10 + 15 = 26 \).

Hence, there are 26 group homomorphisms from \( \mathbb{Z}_2 \) to \( S_5 \). The correct answer is 26.