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Let $G$ be the group $\{\pm1, \pm i \}$ with multiplication of complex numbers as composition. Let $H$ be the quotient group $\mathbb{Z}/4 \mathbb{Z}$. Then the number of nontrivial group homomorphisms from $H$ to $G$ is

1. $4$
2. $1$
3. $2$
4. $3$

recategorized | 20 views

Hi, first of all notice that G is cyclic and has four elements (it is generated by i). Therefore it is isomorphic to Z/4Z and then the number of non trivial group homomorphisms from H to G will equal the number of non trivial group homomorphisms from H = Z/4Z to itself. Now, a group homomorphism is between cyclic groups is uniquely determined by the image of a generator, so since Z/4Z is generated by [1] you see that there are 4 group homomorphisms:
- f([1]) = [0]
- f([1]) = [1]
- f([1]) = [2]
- f([1]) = [3],
where the first one is trivial (it is the 0-map), while the others are not (well, actually the second one is also trivial since it is the identity, but when you replace Z/4Z with G it becomes the map sending [1] to i, which is no longer the identity). Therefore you have 3 non trivial group homomorphisms from H to G.
I hope this helped you. Bye!
by (17 points)