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Let $G$ be a finite abelian group generated by two elements $a$ and $b$. I am trying to prove that $G$ is isomorphic to the direct product of two cyclic group $C_r$ and $C_s$, where the value of $r$ and $s$ depend on $|a|$, $|b|, d = |\langle a \rangle \cap \langle b \rangle |.$ I have defined a map $$ \psi : \frac{\langle a \rangle}{\langle a \rangle \cap \langle b \rangle } \times \langle b \rangle : \rightarrow \langle a, b \rangle $$

such that $\psi(\bar{a^i}, b^j) = (a^i b^j).$ But this map is not well defined. Can we defined another map or how to prove that $G$ is isomorphic to the direct product of two cyclic group $C_r$ and $C_s$, where the value of $r$ and $s$ depend on $|a|$, $|b|, d = |\langle a \rangle \cap \langle b \rangle |.$ Any help would be appreciated. Thank you.

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    $\begingroup$ There's something to be wary of in saying a finite abelian group is "generated by two eleroup ments." While you clearly require both elements to have finite orders, unless those orders are coprime, the group they generate is not uniquely determined. Here you deal with this by asking also for the order $d$ of the intersection of the two cyclic subgroups. $\endgroup$ – hardmath Apr 21 '18 at 19:01
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Here's a different approach:

Theorem: Let $A$ be an abelian group and suppose $A$ is generated by $a, b$, where $a$ and $b$ have finite order. Then $A$ is isomorphic to the product of two cyclic groups.

Proof: Take generators $a$ and $b$ of $A$, and write $C = \langle a \rangle \subset A$. $A /C$ is cyclic with generator $c = \overline{b}$. There is a canonical quotient map $\pi : A \rightarrow A/C$. $\pi (b) = c$. $c$ is a generator of $A/C$. The order of $b$ is divisible by the order of $A/C$. To see this, note that $\pi$ induces a surjective map $\pi' : \langle b \rangle \rightarrow A/C$, so that $|A/C| \cdot |\ker(\pi')| = \text{ord}(b)$. Write $\text{ord}(b) = n$ and $|A/C| = m$, and take $k \in \mathbb{N}$ such that $mk = n$. $b^k$ then has order $m$ and $\phi(b^k)$ is a generator of $A/C$ (why?).

We show that $\langle b^k \rangle \cap \langle a \rangle = \{ e \}$ and $\langle b^k \rangle \langle a \rangle = A$. Suppose $b^{ki} = a^j$ for some $i, j \in \mathbb{N}$. Applying $\pi$, we see that $c^i = e$, so that $i$ is divisible by $m$. Then $ki$ is divisible by $n$, so that $b^{ki} = e$. Then $b^{ki} = a^j = e$. So $\langle b^k \rangle \cap \langle a \rangle = \{ e \}$. To see that $\langle b^k \rangle \langle a \rangle = A$, take $x \in A$, and take $i$ such that $\pi (x) = c^i$. Then $xb^{-ki} = c^i c^{-i} = e$. We can write $xb^{-ik} = a^j$ since $xb^{-ik} \in \ker(\pi)$. Then $x = a^j b^{ik}$.

It follows from a well known characterization of direct products that $A \cong \langle b^k \rangle \times \langle a \rangle$. Therefore $A$ is the product of two cyclic groups.

Note: Here's some intuition concerning direct products that might help. Suppose we have a quotient map $\pi: A \rightarrow B$. When is $A$ isomorphic to the direct product of $B$ and $\ker(\pi)$? It turns out that this is true whenever there is another map $\iota : B \rightarrow A$ such that $\pi \circ \iota = \text{Id}_B$. It is a good exercise to verify this (and it's similar to the proof above). In our case, we created such a situation as this by taking $\langle a \rangle = \ker(\pi)$, $B = A / \langle a \rangle$, and $\pi : A \rightarrow B$ the canonical quotient map. The rest of what we showed was actually equivalent to showing that we have a map $\iota : B \rightarrow A$ ($\iota : B \rightarrow A$ would send $c$ to $b^k$).

Note: This is actually a specific case of a much more general theorem. All finitely generated abelian groups have a decomposition into a product of cyclic groups (note $\mathbb{Z}$ is cyclic). You can read about the most general formulation of this, in terms of finitely generated modules over a PID, here.

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  • $\begingroup$ You have chosen $k \in \mathbb N$ such that $mk = n$, then ord$(b^k) = \frac{\text{ord}(b)}{g.c.d (\text{ord}(b), k)} = \frac{n}{g.c.d (n, k)} = \frac{n}{g.c.d (mk, k)} = \frac{n}{k} = m$ $\endgroup$ – Struggler Apr 22 '18 at 9:00
  • $\begingroup$ Why $\pi(b^k) = c$ $\endgroup$ – Struggler Apr 22 '18 at 10:50
  • $\begingroup$ You were right, I meant $m$ and not $n$. $\endgroup$ – Dean Young Apr 22 '18 at 16:16
  • $\begingroup$ But how to show $\pi(b^k) = c$ $\endgroup$ – Struggler Apr 22 '18 at 16:28
  • $\begingroup$ Consider $G = \mathbb Z_{12}\times \mathbb Z_2, b = (1, 0), \; a = (4, 1), \; n = 12, \; m = 4, \; k = 3.$. We have $\pi(b )= c = (0, 1)\langle a \rangle$ and $\pi(b^3 )= c^3 \neq c$. Please look at this example. $\endgroup$ – Struggler Apr 24 '18 at 7:06

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