Question

List the Sylow 2-subgroups, Sylow 3-subgroups, and Sylow 5-subgroups of ${\mathrm{\mathbb{Z}}}_{\mathbf{12}}\mathbf{\times }{\mathbf{\mathbb{Z}}}_{\mathbf{12}}\mathbf{\times }{\mathbf{\mathbb{Z}}}_{\mathbf{10}}$. [Section 9.2 is a prerequisite for this exercise]

Short answer

The Sylows 2-sugroups are $\left\{\left(\left[3\times \right],\left[3y\right],\left[3z\right]\right)\in {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{10}\overline{)xyz\in \mathrm{\mathbb{Z}}}\right\}$, the Sylows 3-sugroups are$\left\{\left(\left[4\times \right],\left[4y\right],\left[0\right]\right)\in {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{10}\overline{)xy\in \mathrm{\mathbb{Z}}}\right\}$and Sylows 5-sugroups are $\left\{\left(\left[0\right],\left[0\right],\left[2z\right]\right)\in {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{12}{\mathrm{\mathbb{Z}}}_{10}\overline{)z\in \mathrm{\mathbb{Z}}}\right\}$.

Step-by-step solution

Referring to Corollary 9.14 (Cauchy’s Theorem)

If Gis a finite group whose order is divisible by a prime p, then G contains an element of orderp.

Finding the Sylow 2-subgroups, Sylow 3-subgroups, and Sylow 5-subgroups of ℤ12×ℤ12×ℤ10

As we know, ${\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{10}={\mathrm{\mathbb{Z}}}_{2^2\times 3}\times {\mathrm{\mathbb{Z}}}_{2^2\times 3}\times {\mathrm{\mathbb{Z}}}_{2\times 5}$.

So, according to Corollary 9.14, the order of ${\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{10}$must be divisible by 2, 3, and 5.

Therefore, for any arbitrary $x,y,z\in \mathrm{\mathbb{Z}}$, the Sylow p-subgroups can be written as:

The Sylows 2-sugroups are $\left\{\left(\left[3\times \right],\left[3y\right],\left[3z\right]\right)\in {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{10}\overline{)xyz\in \mathrm{\mathbb{Z}}}\right\}$.

The Sylows 3-sugroups are $\left\{\left(\left[4\times \right],\left[4y\right],\left[0\right]\right)\in {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{10}\overline{)xy\in \mathrm{\mathbb{Z}}}\right\}$.

The Sylows 5-sugroups are $\left\{\left(\left[0\right],\left[0\right],\left[2z\right]\right)\in {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{10}\overline{)z\in \mathrm{\mathbb{Z}}}\right\}$.

Hence, the Sylows 2-sugroups are $\left\{\left(\left[3\times \right],\left[3y\right],\left[3z\right]\right)\in {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{10}\overline{)xyz\in \mathrm{\mathbb{Z}}}\right\}$, the Sylows 3-sugroups are $\left\{\left(\left[4\times \right],\left[4y\right],\left[0\right]\right)\in {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{10}\overline{)xy\in \mathrm{\mathbb{Z}}}\right\}$, and Sylows 5-sugroups are $\left\{\left(\left[0\right],\left[0\right],\left[2z\right]\right)\in {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{12}\times {\mathrm{\mathbb{Z}}}_{10}\overline{)z\in \mathrm{\mathbb{Z}}}\right\}$.