Question

Write${\mathrm{\mathbb{Z}}}_{30}$ in three different ways as a direct sum of two or more of its subgroups. [Hint: Theorem 9.3.]

Short answer

The ${\mathrm{\mathbb{Z}}}_{30}$ can be written as, ${\mathrm{\mathbb{Z}}}_{30}=M\oplus N$, ${\mathrm{\mathbb{Z}}}_{30}=M\oplus N_1\oplus N_2$, and ${\mathrm{\mathbb{Z}}}_{30}=M\text{'}\oplus N_2$.

Step-by-step solution

Subgroups of a group

If A and B are normal subgroups of a group G such that $G=AB$and , then $G=A\times B$.

Write ℤ30 in three different ways as a direct sum of two or more of its subgroups

Every subgroup of${\mathrm{\mathbb{Z}}}_{30}$ will be normal subgroup because${\mathrm{\mathbb{Z}}}_{30}$ is abelian.

Consider two subgroups $M=\left\{0,15\right\}$and

$N=\left\{0,2,4,6,8,10,12,14,16,18,20,22,24,26,28\right\}$.

Then $M\cap N=\left\{0\right\}$.

The direct sum for above two subgroups for${\mathrm{\mathbb{Z}}}_{30}$ will be ${\mathrm{\mathbb{Z}}}_{30}=M\oplus N$.

The subgroup N can be further written as sum of two subgroups $N_1=\left\{0,10,20\right\}$, and $N_2=\left\{0,6,12,18,24\right\}$.

The intersection of these subgroups is $N_1\cap N_2=\left\{0\right\}$.

The direct sum for the above two subgroups for${\mathrm{\mathbb{Z}}}_{30}$ will be $N=N_1\oplus N_2$. Therefore, ${\mathrm{\mathbb{Z}}}_{30}=M\oplus N_1\oplus N_2$.

A new subgroup M' can further written as sum of two subgroups M and$N_1$ as:

$\begin{array}{rcl}M\text{'}& =& M\oplus N\\ & =& \left\{0,5,10,15,20,25\right\}\end{array}$

So, by theorem 9.3, ${\mathrm{\mathbb{Z}}}_{30}=M\text{'}\oplus N_2$.

Hence, ${\mathrm{\mathbb{Z}}}_{30}$can be written as,${\mathrm{\mathbb{Z}}}_{30}=M\oplus N,{\mathrm{\mathbb{Z}}}_{30}=M\oplus N_1\oplus N_2$ and ${\mathrm{\mathbb{Z}}}_{30}=M\text{'}\oplus N_2$.