Find isomorphisms between two small finite groups $G, H$ of equal cardinality. 
Let $G=\{2,4,8,10,14,16\}\pmod{18}$ and $H=\{3,6,9,12,15,18\}\pmod{21}.$

Show that, under integer multiplication mod $18$ and mod $21$ respectively, each of these forms a group.
Establish isomorphism $between the two groups, and
Find the number of isomorphisms possible from set $G\to H$.
Need check if the two are groups by checking the property of Closure and inverse.
Inverse candidate for
$G=\{2,4,8,10,14,16\}\pmod{18}=\{ 14,16,8,10,2,4\}=G$,
with identity $e=10$.
$2^{-1}=14, 4^{-1}= 16, 8^{-1}=8, 10^{-1}= 10, 14^{-1}=2, 16^{-1}= 4.$
Inverse candidate for
$H=\{3,6,9,12,15,18\}\pmod{21}=\{ 12,6,18,3,15,9\}= H$,
with identity $e=15$.
$3^{-1}=12, 6^{-1}= 6, 9^{-1}=18, 12^{-1}= 3, 15^{-1}=15, 18^{-1}= 9.$

Need find order of each element to find mapping under $H\mapsto G$.
\begin{array}{c|cccc}
g & 3&6& 9& 12&15& 18\\ \hline
\varphi(g) &&8 & &&10 &\\\hline
\end{array}
Seems all elements except two in each group have order $=6$ in each group.
First one is identity.
In $G $, $|8|=2$ and in $H$ have $|6|=2.$
$\langle 2\rangle =\{2, 4,8,16, 14, 10\},$ same for others too.
$\langle 4\rangle =\{4, 16, 10, 8, 14\}.$
$\langle 8\rangle =\{8, 10\},$
$\langle 14\rangle =\{14, 16, 8, 4, 2,10 \},$
Identity has order$=1$. Say in group $G$, have: $10, 100\equiv 10\pmod {18}$.
Confused.

For last part, have no idea, except that $4$ non-identity elements in each set can map to $4$ such in the other set. So, $16$ Isomorphisms.
What should be table entries in part (b)?
 A: Since there are six elements in each underlying set and $6=2\times 3$, we can use the following theorem:

Theorem: Let $p$ be an odd prime. Then each group of order $2p$ is isomorphic to either $\Bbb Z_{2p}$ or $D_p$.

For a proof, see Theorem 7.3 of Gallian's, "Contemporary Abstract Algebra (Eighth Edition)".
Only one of $\Bbb Z_6$ and $D_3$ is abelian, and multiplication is commutative; therefore, if they are groups, we must have $G\cong \Bbb Z_6\cong H$.
But by inspection of the multiplication tables, we have
$$\begin{align}
2^1&=2,\\
2^2&=4,\\
2^3&=8,\\
2^4&=16,\\
2^5&=16\times 2=32\equiv 14,\\
2^6&=2^5\times 2\equiv 14\times 2=28\equiv 10,
\end{align}$$
and $10$ is the identity; thus $G\cong\langle 2\rangle$.
Similarly, $H\cong \langle 3\rangle$ with identity $15$.
Isomorphisms of cyclic groups always send generators to generators; thus
$$\begin{align}
\varphi: G&\to H,\\
2&\mapsto 3
\end{align}$$
defines the isomorphism $\varphi(2^n)=3^n.$
The abstract group $\Bbb Z_6\cong\langle a\mid a^6\rangle$ has automorphism group $\Bbb Z_2$, which we can see as $a$ and $a^5$ are generators, again sent to generators, meaning there are two isomorphisms between $G$ and $H$. Can you write them explicitly?

 We have $\varphi$ above and the isomorphism given by $$\begin{align} \theta: G&\to H,\\ 2&\mapsto 3^5\equiv 12. \end{align}$$

A: Note $G$ and $H$ are cyclic groups with order $6$
For $G$, $e=10$, $G=\langle 2\rangle$, $~~~2^1=2,~2^2=4,~2^3=8,~2^4=16,~2^5=14,~2^6=10=e$
For $H$, $e=15$, $H=\langle 3\rangle$, $~~~3^1=3,~3^2=9,~3^3=6,~3^4=18,~3^5=12,~3^6=15=e$
Isomorphism: $\phi(2^k)=3^k$
Since finite cyclic group with order $n$ is isomorphic to $\mathbb{Z}_n$, both $G$ and $H$ are isomorphic to  $\mathbb{Z}_6$.
For cyclic group, the isomorphism has to map the generator to generator, so consider how many generators in the group.
For $\mathbb{Z}_6$, there are two generators, which are $1$ and $5$, so there are two isomorphisms.
