Find isomorphisms between two small finite groups $G, H$ of equal cardinality.

Question

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 $e=15$. $3^{-1}=12, 6^{-1}= 6, 9^{-1}=18, 12^{-1}= 3, 15^{-1}=15, 18^{-1}= 9.$

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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.

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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)?

Answer 1

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.

Answer 2

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 $$\begin{align} \theta: G&\to H,\\ 2&\mapsto 3^5\equiv 12. \end{align}$$