How many different homomorphisms $\varphi : \mathbb{Z}_{18} \to \mathbb{Z}_{24}$ so $|Im\varphi| =6$

Trying to solve the following question:

Let $$a$$ be the number of different homomorphisms of group $$\varphi : \mathbb{Z}_{18} \to \mathbb{Z}_{24}$$ so $$|Im\varphi| =6$$.

Find $$a$$.

In the solution they stated that $$\mathbb{Z}_{24}$$ is cyclic and it has only one subgroup of order $$6$$. Than the number of different homomorphism is based on the number of generators $$b$$ of that subgroup because $$\varphi(1)=b$$ when $$b$$ is the generator of a subgroup of order $$6$$. There are two $$b$$'s so $$a=2$$.

I don't understand the solution. Why due to $$|Im\varphi|=6$$ we need to talk about subgroups of that order? Also, why $$\varphi(1)=b$$ and why there are two?

A slight variation. There is only one subgroup of $$\mathbb Z/24$$ of order $$6$$, and it is cyclic, so it is isomorphic to $$\mathbb Z/6$$. The question now is how many surjective homomorphisms are there $$\mathbb Z/18 \to \mathbb Z/6$$. Such a homomorphism is determined by the image of the generator. Thus, the question is now, how many generators of $$\mathbb Z/6$$ are there? The number of generators is the number of elements in $$1,2,3,4,5$$ that are relatively prime to $$6$$, i.e., $$1,5$$, thus the answer is $$a=2$$.

This is due to the following facts, that you should be able to verify yourself:

• the image of a group homomorphism is a group

• since $$\mathbb Z_{18}$$ is cyclic and generated by 1, it's image is cyclic generated by $$\varphi(1)$$

• as you mention, the only subgroup of $$\mathbb Z_{24}$$ of order $$6$$ is $$\{0,4,8,12,16,20\}$$. The generators of this group are those elements of order 6, namely $$4$$ and $$20$$. So you can have one $$\varphi_1$$ given by $$\varphi_1(1)=4$$, and another one given by $$\varphi_2(1)=20$$.

Because if $$\varphi :\mathbb Z_{12}\to \mathbb Z_{18}$$ is a Group homomorphism, then $$\text{Im}(\varphi )$$ is a subgroup of $$\mathbb Z_{24}$$. So if $$|\text{Im}(\varphi )|=6$$ then $$\text{Im}(\varphi )=4\mathbb Z/24\mathbb Z$$ (because it's the only subgroup of $$\mathbb Z_{24}$$ of order $$6$$). Now, $$\varphi$$ is uniquely determinated by $$\varphi (1)$$ (because $$\mathbb Z_{18}$$ is cyclic). Since $$4\mathbb Z/24\mathbb Z=\{0,4,8,12,16,20\}=\left<4\right>=\left<20\right>,$$ where I abusively denote $$a$$ the class of $$a$$, then any $$\varphi (1)\in \{0,4,8,12,16,20\}$$ will denote an homomorphism. Now you need to have that $$\varphi (1)$$ has order $$6$$ (because $$\text{Im}(\varphi )$$ is a cyclic group of order $$6$$). The only element of order $$6$$ are $$4$$ and $$20$$. So, the only non trivial homomorphisms are determined by $$\varphi (1)=4$$ and $$\varphi (1)=20$$.

• why $4$ and $20$ are of order $6$? $4^6 \mod 24 = 16\neq 1$ – vesii Jul 10 '19 at 18:11
• Because $6\cdot 4=0$ and $k\cdot 4\neq 0$ for all $k\in \{1,...,5\}$. Same for $20$. – Surb Jul 10 '19 at 19:05
• additive order, not multiplicative order. – Justin Young Jul 11 '19 at 17:20