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In the algebra lecture i need to solve the following exercise

Use the universal property of the tensor product to show that $$(\Bbb{Z}/4\Bbb{Z}) \otimes (\Bbb{Z}/6\Bbb{Z})\cong \Bbb{Z}/2\Bbb{Z}$$

Here is my job:

I tried to follow the example seen in the lecture where we showed that $(\Bbb{Z}/2\Bbb{Z}) \otimes (\Bbb{Z}/3\Bbb{Z})=0$. In that example the professor listed all the generators of $(\Bbb{Z}/2\Bbb{Z}) \otimes (\Bbb{Z}/3\Bbb{Z})$ and then showed that every generator is actually zero just by using the properties of the tensors. Namely

$$\begin{array}{|cc|}\hline 0\otimes n = 0\otimes n + 0\otimes n &\implies 0\otimes n =0 \\\hline 1\otimes0=1\otimes 0 + 1 \otimes 0 &\implies 1\otimes 0 =0 \\\hline 1\otimes 2 =2(1\otimes 1) = 0\otimes1&\implies1\otimes 2=0 \\\hline 1\otimes1 =1\otimes 4 = 1\otimes2 + 1\otimes 2 & \implies 1\otimes 1 =0\\\hline \end{array}$$

for $n= 0,1,2$.

Now to show that $(\Bbb{Z}/4\Bbb{Z}) \otimes (\Bbb{Z}/6\Bbb{Z})\cong \Bbb{Z}/2\Bbb{Z}$ I tried to list all the generator and try to show that they are equal to $0$ or $1$. We have $4\cdot 6=24$ generators and i noticed that if a generator has a even integer than it is equal $0$. The problem is that i can not show that some generators are equal $1$. Indeed the generators equal $0$ are

$$\begin{array}{|cc|}\hline 0\otimes n=0\otimes n + 0\otimes n& \implies 0\otimes n =0 \,\,\,n\in \{0,1,2,3,4,5\} \\\hline n\otimes 0 = n\otimes 0 + n \otimes 0&\implies n \otimes 0 =0 \,\,\, n \in \{0,1,2,3\}\\\hline \end{array}$$

and \begin{align*} 1\otimes 2=0 && 1\otimes 4=0 && 1\otimes 6=0 \\ 2\otimes n=0 && 3\otimes 2=0 && 3\otimes 4=0 \\ 3\otimes 6=0 \end{align*} for $n \in \{0,1,2,3,4,5\}$

The other generator can be brought back at the case $1\otimes 1$ and therefore i wish i could show $1\otimes 1$ can be considered as $1$, but i'm not able to. I know that i should use the universal property of the tensor product but i don't really know how to use it, therefore i think that there must be another procedure to solve the exercise. Is what i've done so far correct?

Bman72
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    Your calculation looks correct, and it shows that $\mathbb{Z}/2 \to \mathbb{Z}/4 \otimes \mathbb{Z}/6$, $1 \mapsto 1 \otimes 1$ is surjective. But with such calculations you won't get injectivity. Anyway, better don't calculate so much and follow the hint with the universal property. Consider $\mathbb{Z}/4 \times \mathbb{Z}/6 \to \mathbb{Z}/2$, $([a],[b]) \mapsto [ab]$, show that this is well-defined and bilinear, in fact a universal bilinear map (alternatively, this map shows you $1 \otimes 1 \neq 0$, which is injectivity from above). For details and generalizations, see the linked question. – Martin Brandenburg Oct 11 '14 at 16:01
  • @martinbrandenburg thank you for your attention and for linking me the link with the general statement. I will look at it ;-) Have a nice evening! – Bman72 Oct 11 '14 at 17:20

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