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Exercise (Practice With Orders In External Direct Products)

In each part below, you are given an element $(g,h)$ and a group $G\oplus H$. Compute the order of $g$ in $G$, the order of $h$ in $H$, and the order of $(g,h)\in G\oplus H$.

  1. $(1,2)\in \mathbb{Z}_2\times\mathbb{Z}_3$
  2. $(1,2)\in \mathbb{Z}_6\times\mathbb{Z}_4$
  3. $(1,2)\in \mathbb{Z}_6\times\mathbb{Z}_8$
  4. $(10,20)\in \mathbb{Z}_{4000}\times\mathbb{Z}_{600}$

Click to see a solution.

  1. We know $|1| = 2$ and $|2|=3$, and we can compute $(1,2)^k$ for each $k$ from $1$ to $6$ to obtain the sequence $((1,2),(0,1),(1,0),(0,2),(1,1),(0,0))$. This shows that $|(1,2)|=6$.
  2. We know $|1| = 6$ and $|2|=2$, and we can compute $(1,2)^k$ for each $k$ from $1$ to $6$ to obtain the sequence $((1,2),(2,0),(3,2),(4,0),(5,2),(0,0))$. This shows that $|(1,2)|=6$.
  3. We know $|1| = 6$ and $|2|=4$. Computing $(1,2)^k$ for each $k$ from $1$ to $12$ will show that the order is 12. Alternately, notice that in the product $(1,2)^k$, the first component will only equal 0 when $k$ is multiple of 6. Similarly, the second component will only be 0 when $k$ is a multiple of $4$. The number 12 is the smallest number for which both components will equal 0.
  4. We know $|10| = 400$ and $|20|=30$. In the product $(10,20)^k$, the first component will only equal 0 when $k$ is multiple of 400, and the second component will only be 0 when $k$ is a multiple of $30$. The least common multiple of 400 and 30 is 1200, which is the order of $(10, 20)$