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Exercise (Practice With Cosets Of $3\mathbb{Z}$)

Let $G=\mathbb{Z}$ and let $H=3\mathbb{Z}$, the multiples of $3$.

  1. List the elements of $0+H$, $1+H$, $2+H$, $3+H$, $4+H$, $5+H$, $17+H$, and $-11+H$.
  2. How many different sets did you get?

Click to see a solution.

We know that $H=\{\ldots,-6, -3, 0, 3, 6, \ldots\}$. Adding 1 to each entry gives $H=\{\ldots,-5, -2, 1, 4, 7, \ldots\}$, which is the same as the set of integers whose remainder is 1 after dividing by 3. Similarly we have $$\begin{align} 0+H &= \{\ldots,-6, -3, 0, 3, 6, \ldots\}, \\ 1+H &= \{\ldots,-5, -2, 1, 4, 7, \ldots\}, \\ 2+H &= \{\ldots,-4, -1, 2, 5, 8, \ldots\}, \\ 3+H &= \{\ldots,-3, 0, 3, 6, 9, \ldots\} = \{\ldots,-6, -3, 0, 3, 6, \ldots\} = 0+H,\\ 4+H &= \{\ldots,-2, 1, 4, 7, 10, \ldots\} = \{\ldots,-5, -2, 1, 4, 7, \ldots\} = 1+H,\text{ and}\\ 5+H &= \{\ldots,-1, 2, 5, 8, 11, \ldots\} = \{\ldots,-4, -1, 2, 5, 8, \ldots\} = 2+H. \end{align}$$ In our computations above, we see that $k+H$ is equivalent to $k\pmod 3+H$. Since $17\pmod 3=2$ and $-11\pmod 3=1$, we have that $17+H = 2+H$ and we also have $-11+H = 1+H$.

There are only three different cosets. The cosets are $0+H$, $1+H$, and $2+H$. Every other coset is equal to one of these.