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Define $C_n$ as the Cyclic group of order $n$. Define $G = ((C_n × C_m), +)$, where the sum is defined such that $a$, $c$ $\epsilon$ $C_n$ and $b$, $d$ $\epsilon$ $C_m$, $(a, b) + (c, d) = ((a + c) \pmod{n},(b + d) \pmod{m})$

(i) Show that $\forall m$, $n$ $\epsilon$ $\mathbb{N}$, $G = ((C_n × C_m), +)$ is a group.

(ii) Let $G = ((C_3 × C_5), +)$ Use the Fundamental Theroem of Cyclic group to find all the generator for $G$, the order of $G$ and the subgroups of $G$.

User203940
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John Perks
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1 Answers1

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The comment already essentially answers this but I'll walk you through it. In order for this to be a group, we need it to satisfy three axioms (which you can find here: https://en.wikipedia.org/wiki/Group_(mathematics)):

(1) Closure -- If we take our group to be $G$, then for all $a, b \in G$ we need $a + b \in G$. This is easy to see.

(2) Associativity -- For all $a, b, c \in G$ we require $a + (b + c) = (a + b) + c$. Let $a, b, c \in (C_n \times C_m)$. Then we have $a = (a_1, a_2)$, $b = (b_1, b_2)$, and $c = (c_1, c_2)$ for $a_1, b_1, c_1 \in C_n$ and $a_2, b_2, c_2 \in C_m$. You then must show that $(a_1, a_2) + ((b_1, b_2) + (c_1, c_2)) = ((a_1, a_2) + (b_1, b_2)) + (c_1, c_2)$.

(3) Identity -- There needs to be an identity element. Select $e = (e_1, e_2)$ where $e_1 \in C_n$ is the identity element and $e_2 \in C_m$ is the identity element. Then you need to show that $a + e = e + a = a$ for all $a = (a_1, a_2) \in (C_n \times C_m)$.

(4) Inverse -- For every element, there is an inverse element. Take some element $a = (a_1, a_2) \in (C_n \times C_m)$. Then show that $(a_1, a_2) + (a_1^{-1}, a_2^{-1})$ will give you the identity element. Define $a^{-1} := (a_1^{-1}, a_2^{-1})$.

Since it satisfies all of our axioms, it must be a group. The second part should follow pretty easily, but let me know if you need help.

User203940
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  • As a hint for the second part, I'll leave this: https://math.stackexchange.com/questions/5969/product-of-two-cyclic-groups-is-cyclic-iff-their-orders-are-co-prime – User203940 Sep 18 '17 at 18:33
  • Can you help me out? Do I have to find the order of $C_3 X C_5 = C_15$ or the order of $((2+1)(mod3),(3+2)(mod5))5 – John Perks Sep 18 '17 at 20:47
  • Notice that 3 and 5 are coprime. Using the link, we have that $C_3 \times C_5$ generates a cyclic group of size 15. Since it's a cyclic group of size 15, what do we know about the generators? Using this https://en.wikipedia.org/wiki/Cyclic_group we also have that $C_{15} \cong \mathbb{Z}/15\mathbb{Z}$. This should help you identify subgroups. – User203940 Sep 19 '17 at 13:27