Recently the theme has been to construct new groups from old groups like the quotient group. There are other ways to construct groups like the following

Definition: Direct Products

Suppose $G$ and $H$ are groups. Then the Direct Product: $$ \begin{align*} G \times H = \{(g,h) \ | \ g \in G, h \in H\} \end{align*} $$ With Operation: $$ \begin{align*} (g_1, h_1) \cdot (g_2, h_2) = (g_1g_2, h_1h_2) \end{align*} $$ is a group. The identity element $e_{G \times H} = (e_G, e_H)$ and the inverse of $(g,h)$ is $(g^{-1},h^{-1})$.

You can verify that this is indeed a group (Exercise).

Example

Let $G = \langle a \rangle$ where $o(a) = 2$ and $H = \langle b \rangle$ with $o(b) = 2$. Then

$$ \begin{align*} G \times H = \{ (e,e), (a,e), (e,b), (a,b)\} \end{align*} $$

Note that this is a group. Since it has four elements then it must be isomorphic to either $V$ ( symmetries of the rectangle) or the cyclic one (so we need an element of order 4). To classify this group, we need to find the order of each element.

$$ \begin{align*} (a,b)^2 &= (a,b) \cdot (a,b) = (aa, bb) = (e, e) \quad \text{(because the order of a and b is 2)} \\ (a,e)^2 &= (e,e) = (e,b)^2. \end{align*} $$

From this we see that each element is of order 2 since $G \times H$ is isomorphic to $V$ (The Klien 4-group which is isomorphic to $\mathbf{Z}_2 \times \mathbf{Z}_2$).

Properties of Direct Product Groups

These are some facts about direct product groups

Proposition

Let $G$ and $H$ be groups. Then the size of $G \times H$ is $|G \times H| = |G||H|$.

Proposition

Let $G_1 \cong G_2$ and $H_1 \cong H_2$ be groups. Then $G_1 \times H_1 \cong G_2 \times H_2$.

Proof
To show that two groups are isomorphic, we need to provide an isomorphism between the two groups. Since we know that $G_1$ and $G_2$ are isomorphic, then let $\alpha : G_1 \rightarrow G_2$ be an isomorphism. Similarly, since $H_1$ and $H_2$ are isomorphic, then let $\beta : H_1 \rightarrow H_2$ be an isomorphism. Then we claim that

$$ \begin{align*} \gamma : G_1 \times H_1 &\rightarrow H_2 \times G_2 \\ \gamma(g, h) &= (\alpha(g), \beta(h)). \end{align*} $$

is an isomorphism. [TODO … verify]

Generalizing Direct Products

We can generalize the definition of direct products to say

Definition: Generalized Direct Products

Suppose $G_1, G_2,...,G_r$ are groups. Then $$ \begin{align*} \bar{G} &= G_1 \times G_2 \times .... \times G_r \\ &= \{(g_1, g_2,...,g_r) \ | \ g_i \in G_i\} \end{align*} $$ With Operation: $$ \begin{align*} (g_1, g_2, ..., g_r) \cdot (g'_1,g'_2, ..., g'_r) = (g_1g'_1, g_2g'_2, ..., g_rg'_r) \end{align*} $$ is a group.

Example

Let’s take the previous example but slightly modify it so that $G = \langle a \rangle$ where $o(a) = 2$ and $H = \langle b \rangle$ with $o(b) = 3$. So now we’re get 6 elements instead in the group

$$ \begin{align*} G \times H = \{ (e,e), (a,e), (e,b), (a,b), (e,b^2), (a, b^2)\} \end{align*} $$

This group is cyclic. It must contain an element of order 5 that generates the group. If we let $(a, b) = x$, then we’ll see that

$$ \begin{align*} (a,b) \cdot (a,b) &= (e, b^2) = x^2 \\ (e, b^2) \cdot (a,b) &= (a, e) = x^3 \\ (a, e) \cdot (a,b) &= (e, b) = x^4 \\ (e, b) \cdot (a,b) &= (a, b^2) = x^5. \end{align*} $$

At some point when we get to $x^4$, we know by Lagrange, that the order of the element will be 6. This group is in fact isomorphic to $\mathbf{Z}_2 \times \mathbf{Z}_3 \cong \mathbf{Z}_6$.

Why was this example different? why was this one cyclic. Because 2 and 3 are prime numbers. $a$ has order $2$ and $b$ is of order $3$. To get to the identity element, we will need a power that is the least common multiple which is 6.

We can formalize this idea in the next proposition

Proposition

Let $a, b \geq 1$. The formula $$ \begin{align*} \gamma(x) = ([x]_a, [x]_b) \end{align*} $$ defines a homomorphism $\gamma: \mathbf{Z} \rightarrow \mathbf{Z}_a \times \mathbf{Z}_b$ such that $$ \begin{align*} \ker(\gamma) = \mathbf{Z}m \ \text{ (multiples of $m$ where } m=lcm(a,b)). \end{align*} $$

Proof
First note that $\gamma$ is well defined. Next, we want to show that $\gamma$ is a homomorphism. Since we’re working on $\mathbf{Z}_a$ and $\mathbf{Z}_b$, then we want to show that $\gamma(x+y) = \gamma(x) + \gamma(y)$. Observe that

$$ \begin{align*} \gamma(x+y) &= ([x+y]_a, [x+y]_b) \\ &= ([x]_a + [y]_a, [x]_b, [y]_b) \\ &= ([x]_a, [x]_b) + ([y]_a, [y]_b) \\ &= \gamma(x) + \gamma(y) \\ \end{align*} $$

Next, we want to show that the kernel is the set of multiples of $m$ so $\ker(\gamma) = \mathbf{Z}m$ where $m = lcm(a,b)$. By definition,

$$ \begin{align*} \ker(\gamma) &= \{x \in \mathbf{Z} \ | \ \gamma(x) = ([x]_a, [x]_b) = ([0]_a, [0]_b)\}. \end{align*} $$

In other words, we want $x$ such that $x \equiv 0 (\bmod a)$ and $x \equiv 0 (\bmod b)$ divides $x$. That is $a$ divides $x$ and $b$ divides $x$. This is the set of multiples of $m$. Therefore

$$ \begin{align*} \ker(\gamma) &= \mathbf{Z}a \cap \mathbf{Z}b = \mathbf{Z}m. \end{align*} $$

where $m = lcm(a,b) = \frac{ab}{gcd(a,b)}$.

A New Homomorphism

As a consequence of this, we will get an injective homomorphism from the quotient group $\mathbf{Z}/\mathbf{Z}m$ which is $\mathbf{Z}_m$ to $\mathbf{Z}_a \times \mathbf{Z}_b$ as follows

$$ \begin{align*} \mathbf{Z}_m &\rightarrow \mathbf{Z}_a \times \mathbf{Z}_b \\ [x]_m &\rightarrow ([x]_a, [x]_b) \end{align*} $$

Why? Recall that we defined a homomorphism $\gamma$ from $\mathbf{Z}$ to $\mathbf{Z}_a \times \mathbf{Z}_b$. Its kernel $K = \mathbf{Z}m$. By the Homomorphism Theorem, we get another homomorphism from the quotient group $\mathbf{Z}/K$ to the same target group $\mathbf{Z}_a \times \mathbf{Z}_b$. Also recall that since $\ker(\gamma) = K = \mathbf{Z}m$, then the new homomorphism is also injective.

When will the homomorphism be surjective? Since it’s injective, then it suffices that to show that $|\mathbf{Z}_m| = |\mathbf{Z}_a \times \mathbf{Z}_b|$. So when will they have the same size?

Theorem

If $a, b \geq 1$ and $gcd(a,b) = 1$. Then we have an isomorphism $$ \begin{align*} \gamma: \mathbf{Z}_m &\rightarrow \mathbf{Z}_a \times \mathbf{Z}_b, \quad (m = ab)\\ [x]_m &\rightarrow ([x]_a, [x]_b) \end{align*} $$

Fact: the reverse, if $gcd(a,b) > 1$, then $\mathbf{Z}_a \times \mathbf{Z}_b \not\cong \mathbf{Z}_{ab}$.

Chinese Remainder Theorem

The previous theorem is in fact the Chinese Remainder Theorem but the Chinese Remainder Theorem is stated in terms of modular arithmetic.

Theorem

If $a, b \geq 1$, $gcd(a,b) = 1$ and $m = ab$. Then for any $\alpha, \beta \in \mathbf{Z}$, there exists $x \in \mathbf{Z}$ such that

$$ \begin{align*} x \equiv \alpha &(\bmod a) \\ x \equiv \beta &(\bmod b) \end{align*} $$

Furthermore, any two such $x$ are congruent modulo $m$.

So now we can generalize our previous theorem to

Theorem

If $a_1, a_2, ... a_r \geq 1$ and $gcd(a_i, a_j) = 1$ if $i \neq j$. Then there is an isomorphism $$ \begin{align*} \mathbf{Z}_m &\rightarrow \mathbf{Z}_{a_1} \times \mathbf{Z}_{a_2} \times ... \mathbf{Z}_{a_r}\\ [x]_m &\rightarrow ([x]_{a_1}, [x]_{a_2},...,[x]_{a_r}) \end{align*} $$ where $m = a_1a_2....a_r$.

For example: $\mathbf{Z}_{240} = \mathbf{Z}_{16} \times \mathbf{Z}_3 \times \mathbf{Z}_5$

The Multiplicative Group

Recall $\mathbf{Z}_n$ is not just a group but a ring with two operations addition and multiplication. Inside this ring, we can focus on the elements that have a multiplicative inverse and form $\Phi(n) = \{u \in \mathbf{Z}_n \ |$ u has a multiplicative inverse $\} \subseteq \mathbf{Z}_n$. $(\Phi(n),\cdot)$ is a group with the multiplication operation. This group is also abelian and is called “modular units”. It is not a subgroup of $\mathbf{Z}_n$.

It turns out that we can use the Chinese Remainder Theorem to decompose $\Phi(n)$, the modular units group. Before listing the theorem recall that if $x \in \mathbf{Z}$. Then $[x]_n \in \Phi(n)$ if and only if $gcd(x,n)=1$.

Theorem

If $a, b \geq 1$, $gcd(a, b) = 1$ and $m = ab$. Then there is an isomorphism $$ \begin{align*} \Phi(m) &\rightarrow \Phi(a) \times \Phi(b) \\ [x]_m &\rightarrow ([x]_{a}, [x]_{b}) \end{align*} $$

Proof
Recall the isomorphism we established earlier

$$ \begin{align*} \gamma: \mathbf{Z}_m &\rightarrow \mathbf{Z}_a \times \mathbf{Z}_b \\ [x]_m &\rightarrow ([x]_{a}, [x]_{b}) \end{align*} $$

This worked with addition. But it is also compatible with multiplication. We didn’t define multiplication on ordered pairs. So define it as follows.

$$ \begin{align*} (u_1, v_1) \cdot (u_2, v_2) = (u_1u_2, v_1v_2) \end{align*} $$

where $u_1, u_2 \in \mathbf{Z}_a$ and $v_1,v_2 \in \mathbf{Z}_b$. The two operations (addition and multiplication) make $R = \mathbf{Z}_a \times \mathbf{Z}_b$ a ring. We have two claims:

Claim 1: $\gamma(xy) = ([xy]_a, [xy]_b)$
This is true because we know $\gamma$ is a homomorphism. So

$$ \begin{align*} \gamma(xy)=\gamma(x)\gamma(y) &= ([x]_a, [x]_b)\cdot ([y]_a, [y]_b) \\ &= ([x]_a[y]_a, [x]_b[y]_b) \\ &= ([xy]_a, [xy]_b). \end{align*} $$

Claim 2: The function $\gamma$ restricts to a function

$$ \begin{align*} \gamma' : \Phi(m) \rightarrow \Phi(a) \times \Phi(b) \end{align*} $$

which is a bijection. why?

if $x \in \mathbf{Z}$ such that $[x]_m \in \Phi(m)$, then we know that $gcd(x,m) = 1$. We defined $m = ab$. So if $x$ is relatively prime to $m$, then it must be relatively prime to the factors of $m$, so $gcd(x,b) = 1$. This means that $[x]_a \in \Phi(a)$ and $[x]_b \in \Phi(b)$.
$\gamma'$ is injective because $\gamma$ is an injective function.
$\gamma'$ is surjective. $\gamma$ is a bijection so every element of $\mathbf{Z}_a \times \mathbf{Z}_b$ has the form $[x]_a, [x]_b$ for some $x \in \mathbf{Z}$. This means that $gcd(x,a) = 1 = gcd(x, b)$. So $gcd(x,ab) = 1$.

Example: $\Phi(30) \cong \Phi(6) \times \Phi(5) \cong \Phi(2) \times \Phi(3) \times \Phi(5)$. Then we can analyze each of these groups. So

$$ \begin{align*} \Phi(5) &\cong \{e\} \\ \Phi(5) &\cong \mathbf{Z}_2 \\ \Phi(5) &\cong \mathbf{Z}_4. \end{align*} $$

References

MATH417 by Charles Rezk
Algebra: Abstract and Concrete by Frederick M. Goodman