MATH 415 Lecture 8

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First Test: October 8 (Tuesday)

Direct Products

(a.k.a. Cartesian product)

Let $S=S_{1}\times S_{2}\times \dots \times S_{n}=\prod _{i=1}^{n}S_{i}$. Then $(s_{1},s_{2},\ldots ,s_{n})\in S$, where $s_{i}\in S_{i}$.

The cardinality of a product is the same as the product of cardinalities

$|S|=|S_{1}|\,|S_{2}|\,\dots \,|S_{n}|$

We are interested in the product of groups. Let $G=G_{1}\times \dots \times G_{n}$, and $g,h\in G$. Let's define our operation as follows:

$g\,h=(g_{1}\,h_{1},g_{2}\,h_{2},\ldots ,g_{n}\,h_{n})$.

Our operation is associative: $(f\,g)\,h=f\,(g\,h)=(f_{1}\,g_{1}\,h_{1},\ldots f_{n}\,g_{n}\,h_{n})$

Our group contains an identity element: $e=(e_{G_{1}},e_{G_{2}},\ldots ,e_{G_{n}})$

All elements of our group are invertable: $g^{-1}=(g_{1}^{-1},g_{2}^{-1},\ldots ,g_{n}^{-1})$.

Alternate Notation: Direct Sums

If the groups of our product are abelian, then we can compute the direct sum of our groups:

$G=G_{1}\oplus G_{2}\oplus \dots \oplus G_{n}$

• $g+h=(g_{1}+h_{1},\ldots ,g_{n}+h_{n})$
• $e=(0,\ldots ,0)$

(note that this is analogous to the direct product of our groups)

The reason we use this is because all abelian groups are isomorphic to an integer group over modular addition.

Example

$\mathbb {Z} _{2}\times \mathbb {Z} _{3}$ can be reexpressed as $\mathbb {Z} _{2}\oplus \mathbb {Z} _{3}$.

The elements of this new group are:

$\left\{(0,0);(0,1);(0,2);(1,0);(1,1);(1,2)\right\}\,\!$

We claim that $\mathbb {Z} _{2}\oplus \mathbb {Z} _{3}\simeq \mathbb {Z} _{6}$.

We can use $(1,1)$ as our generator for all elements of the group, and the justification follows because there is only one cyclic group of order $n=6$, and it is isomorphic to $\mathbb {Z} _{6}$.

Now if we take $\mathbb {Z} _{2}\oplus \mathbb {Z} _{2}$ or $\mathbb {Z} _{3}\oplus \mathbb {Z} _{3}$, these direct sums are not cyclic because there are no generators that will give all elements.

In the first case, $\mathbb {Z} _{2}\oplus \mathbb {Z} _{2}=V$ is called a Klein group.

Theorem 11.5

Theorem. The group $\mathbb {Z} _{m}\times \mathbb {Z} _{n}$ is cyclic and is isomorphic to $\mathbb {Z} _{m\,n}$ if and only if $m$ and $n$ are relatively prime.

Proof. First assume $\gcd {(m,n)}=1$. Then $a=(1,1)\in \mathbb {Z} _{m}\oplus \mathbb {Z} _{n}$ is a generator. We know that the order of $a$ is the minimum $k$ such that $k\,a=e=(0,0)$. $k\,a=(k,k)=(0,0)$, so

${\begin{aligned}k&\equiv 0{\pmod {m}}\\k&\equiv 0{\pmod {n}}\\\end{aligned}}$

From number theory, we know that if $\gcd {(m,n)}=1$, then the smallest $k$ satisfying the equivalences above is $k=m\,n$. This is good because if the order of $(1,1)$ is $m\,n$, then it generates the whole group $\mathbb {Z} _{m}\times \mathbb {Z} _{n}$

Assume $\gcd {(m,n)}=d$ for $d>1$, then ${\frac {m\,n}{d}}$ is divisible by both $m$ and $n$. Thus for any claimed generator $a=(r,s)\in \mathbb {Z} _{m}\times \mathbb {Z} _{n}$, we have $k={\frac {m\,n}{d}}$. So $k(r,s)=\left({\frac {m\,n}{d}}\,r,{\frac {m\,n}{d}}\,s\right)$, where both components are equivalent to 0 mod $m$ and $n$, respectively. Therefore the order of $a$ must be at most ${\frac {m\,n}{d}}$ and cannot generate the whole group.

Corollary. The group $\mathbb {Z} _{m_{1}}\times \mathbb {Z} _{m_{2}}\times \dots \times \mathbb {Z} _{m_{n}}$ is cyclic and isomorphic to $\mathbb {Z} _{m_{1}\,m_{2}\,\dots \,m_{n}}$ if and only if $\gcd {(m_{i},m_{j})}=1$ for all $i$ and $j$ such that $i\neq j$.

Example

Let $(p_{1})^{n_{1}}\,(p_{2})^{n_{2}}\,\dots \,(p_{r})^{n_{r}}$ be the prime factorization of an integer $n$. Then $\mathbb {Z} _{n}\simeq \mathbb {Z} _{(p_{1})^{n_{1}}}\times \dots \times \mathbb {Z} _{(p_{r})^{n_{r}}}$.

In a specific case, $\mathbb {Z} _{72}\simeq \mathbb {Z} _{2^{3}}\times \mathbb {Z} _{3^{2}}=\mathbb {Z} _{8}\times \mathbb {Z} _{9}$ because $72=2^{3}\cdot 3^{2}$.

Theorem 11.9

Let $(a_{1},a_{2},\ldots ,a_{n})\in \prod _{i=1}^{n}G_{i}$, and let $a_{i}$ be of finite order $r_{i}$ in $G_{i}$. Then the order of $(a_{1},\ldots ,a_{n})\in \prod _{i=1}^{n}G_{1}$ is equal to the least common multiple of $r_{1},\ldots ,r_{n}$.

In other words, we are looking for a $k$ such that

$a^{k}=(a_{1}^{k},a_{2}^{k},\ldots ,a_{n}^{k})=(e_{1},e_{2},\ldots ,e_{n})\,\!$

The theorem above claims that $k$ is the least common multiple of the orders of the contained elements.

Proof. We have $r_{i}\mid k$ and $a_{i}^{k}=e_{i}$ in $G_{i}$.

Example

What is the order of $(8,4,10)\in \mathbb {Z} _{12}\times \mathbb {Z} _{60}\times \mathbb {Z} _{24}$

• $8\in \mathbb {Z} _{12}$ has order ${\frac {12}{\gcd {(8,12)}}}={\frac {12}{4}}=3$
• $4\in \mathbb {Z} _{60}$ has order ${\frac {60}{\gcd {(4,60)}}}={\frac {60}{4}}=15$
• $10\in \mathbb {Z} _{24}$ has order ${\frac {24}{\gcd {(10,24)}}}={\frac {24}{2}}=12$

Therefore the order of $(8,4,10)$ is Failed to parse (unknown function "\lcm"): {\displaystyle \lcm{(3,15,12)} = 60} .

Internal Direct Product

Let ${\overline {G}}_{i}=\left\{\left(e_{1},\ldots ,e_{i-1},a_{i},e_{i+1},\ldots ,e_{n}\right)~\mid ~a_{i}\in G_{i}\right\}$ be a subset of $G=\prod _{i=0}^{n}G_{i}$.

${\overline {G}}_{i}$ is a subgroup of $G$.

Furthermore, ${\overline {G}}_{i}$ is isomorphic to $G_{i}$ with $\phi (e_{1},\ldots ,e_{i-1},a_{i},e_{i+1},\ldots ,e_{n})=a_{i}$.

We call $G=\prod _{i=1}^{n}{\overline {G}}_{i}$ the internal direct product, and $G=\prod _{i=1}^{n}G_{i}$ the external direct product.