Math4303 Modern Algebra (Lecture 12)

For the forward direction, use $\mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}=\mathbb{Z}_{n_1n_2}$. if $n_1, n_2$ are coprime.

For the backward, suppose to the contrary that for example $\gcd(n_1,n_2)=d>1$, then $G=\mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}\times H$, where any element in $H$ has order $\leq |H|$ and any element in $\mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}$ has order $<\frac{n_1n_2}{d}$, therefore, all the elements in $G$ will have order strictly less than the size $n_1n_2\ldots n_k$ of the group.

$$\mathbb{Z}_{8}\times\mathbb{Z}_8\times \mathbb{Z}_12$$

the order for $(1,1,1)$ is 24.

What is the maximum order of an element in this group?

Guess:

$8*3=24$

If $G$ is abelian of size $8$, then $G$ is isomorphic to one of the following:

• $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$ (non cyclic)
• $\mathbb{Z}_2\times \mathbb{Z}_4$ (non cyclic)
• $\mathbb{Z}_2$ (cyclic)

And any two of them are not isomorphic

---

Find all abelian group of order $72$.

Since $72=2^3*3^2$, There are 3 possibilities for the $2^3$ part, and there are 2 possibilities for the $3^2$ part.

Note that $\mathbb{Z}_8\times\mathbb{Z}_9$, where $8,9$ are coprime, $\mathbb{Z}_8\times\mathbb{Z}_9=\mathbb{Z}_{72}$, is cyclic.

There are 6 possibilities in total.

Write $G=\mathbb{Z}_{p_1}^{n_1}\times \mathbb{Z}_{p_2}^{n_2}\times \cdots \times \mathbb{Z}_{p_k}^{n_k}$ where $p_i$ are distinct primes.

Therefore $n=p_1^{m_1}\ldots p_k^{m_k}$.

For any divisor $d$ of $n$, we can write $d=p_1^{m_1}\ldots p_k^{m_k}$, where $m_i\leq n_i$.

Now for each $p_i$, we choose the subgroup $H_i$ of size $p_i^{m_i}$ in $\mathbb{Z}_{p_i}^{n_i}$. (recall that every cyclic group of size $r$ and any divisor $s$ of $r$, there is a subgroup of order $s$. If the group is generated by $a$, then use $a^{\frac{r}{s}}$ to generate the subgroup.)

We can construct the subgroup $H=H_1\times H_2\times \cdots \times H_k$ is the subgroup of $G$ of order $d$.

Consider $G=S_3$