Suppose that $\phi$ is a homomorphism from $\mathbb{Z}_{36}$ to a group of order $24$. Find all homomorphic images and for each images determine $\ker(\phi)$.
$\underline{\text{Solution:}}$
We know homomorphism completely determined by the generators of domain.
$\underline{\text{Possible orders for }\phi(1):}$
$1\in\mathbb{Z}_{36}$, $\vert 1 \vert=36$. Also we know in a homomorphism $\mathrm{o}(\phi(1))\vert \mathrm{o}(1)=36$, so $\mathrm{o}(\phi(1))\vert 36$. And by Lagrange Theorem, $\mathrm{o}(\phi(1))\vert 24$. Hence, $$\mathrm{o}(\phi(1))=1,2,3,4,6,12$$
We know give a property of group homomorphism:
$$\text{Let G be a group, H be a subgroup of G and } \phi \text{ be a homomorphism. If H is cyclic then } \phi(H) \text{ also cyclic.}$$ $$\text{Every finite cyclic group with order n is isomorphic to the integers modulo n}$$
$\underline{\text{Structure of } \phi(1):}$
Using this theorem since $\langle 1 \rangle < \mathbb{Z}_{36}$ we know $\langle\phi(1)\rangle$ is cyclic. So we can write:
$$\mathrm{o}(\phi(1))=1, \; \phi(1)=\lbrace e \rbrace \cong {e}\\ \mathrm{o}(\phi(1))=2, \; \phi(1)=\lbrace e, a \rbrace \cong \mathbb{Z}_{2} \\ \mathrm{o}(\phi(1))=3, \; \phi(1)=\lbrace e,a,a^2 \rbrace \cong \mathbb{Z}_{3} \\ \mathrm{o}(\phi(1))=4, \; \phi(1)=\lbrace e,a,a^2,a^3 \rbrace \cong \mathbb{Z}_{4}\\ \mathrm{o}(\phi(1))=6, \; \phi(1)=\lbrace e,a,a^2,...,a^5 \rbrace \cong \mathbb{Z}_{6}\\ \mathrm{o}(\phi(1))=12, \; \phi(1)=\lbrace e,a,a^2,a^3,...,a^{11} \rbrace \cong \mathbb{Z}_{12}\\$$
Now, we investigate for each groups we need to find kernels:
$\underline{\text{Finding kernels to the corresponding images}:}$
We know $\ker(\phi) < \mathbb{Z}_{36}$ we will use this.
$\underline{\text{For } \phi(1) \cong \mathbb{Z}_{12}:}$
$$\dfrac{\mathbb{Z}_{36}}{\ker(\phi)} \cong \mathbb{Z}_{12} \rightarrow \mathrm{o}(\ker(\phi))=3 \text{ and} \\ \ker(\phi) < \mathbb{Z}_{36}, \; \left\langle \frac{36}{3}\right\rangle=\langle 12\rangle =\lbrace 0,12,24 \rbrace$$
$\underline{\text{For } \phi(1) \cong \mathbb{Z}_{6}:}$
$$\dfrac{\mathbb{Z}_{36}}{\ker(\phi)} \cong \mathbb{Z}_{6} \rightarrow \mathrm{o}(\ker(\phi))=6 \text{ and} \\ \ker(\phi) < \mathbb{Z}_{36}, \; \left\langle \frac{36}{6}\right\rangle=\langle 6\rangle =\lbrace 0,6,12,18,24,30 \rbrace$$
$\underline{\text{For } \phi(1) \cong \mathbb{Z}_{4}:}$
$$\dfrac{\mathbb{Z}_{36}}{\ker(\phi)} \cong \mathbb{Z}_{4} \rightarrow \mathrm{o}(\ker(\phi))=9 \text{ and} \\ \ker(\phi) < \mathbb{Z}_{36}, \; \left\langle \frac{36}{9}\right\rangle=\langle 4\rangle =\lbrace 0,4,8,12,...,32 \rbrace$$
$\underline{\text{For } \phi(1) \cong \mathbb{Z}_{3}:}$
$$\dfrac{\mathbb{Z}_{36}}{\ker(\phi)} \cong \mathbb{Z}_{3} \rightarrow \mathrm{o}(\ker(\phi))=12 \text{ and} \\ \ker(\phi) < \mathbb{Z}_{36}, \; \left\langle \frac{36}{12}\right\rangle=\langle 3\rangle =\lbrace 0,3,6,9,12,...,33 \rbrace$$
$\underline{\text{For } \phi(1) \cong \mathbb{Z}_{2}:}$
$$\dfrac{\mathbb{Z}_{36}}{Ker(\phi)} \cong \mathbb{Z}_{2} \rightarrow \mathrm{o}(\ker(\phi))=18 \text{ and} \\ \ker(\phi) < \mathbb{Z}_{36}, \; \left\langle \frac{36}{18}\right\rangle=\langle 2\rangle =\lbrace 0,2,4,...,34 \rbrace$$
$\underline{\text{For } \phi(1) \cong \lbrace e \rbrace:}$
$$\dfrac{\mathbb{Z}_{36}}{\ker(\phi)} \cong \lbrace e \rbrace \rightarrow \mathrm{o}(\ker(\phi))=36 \text{ and} \\\ker(\phi) < \mathbb{Z}_{36}, \; \left\langle \frac{36}{36}\right\rangle=\langle 1\rangle =\lbrace 0,1,2,...,35 \rbrace$$
Are there any mistake? Thanks in advance!