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Okay, so I'm working on a homework problem in abstract algebra, and I have found the solution already, what I want to know is why my initial line of reasoning didn't work - i..e, what have I done or assumed that I am not allowed to do.

The problem is as follows. Suppose that $\phi: Z_{50} \rightarrow Z_{15}$ is a homomorphism of groups, such that $\phi (7)=6$ Determine $\phi (x)$ for every $x \in Z_{50}$. Here was my line of reasoning.

$ \phi (7)=6 \implies [\phi (7)]^2=6^2 \implies \phi(49)=6$ by the properties of a homomorphism (which I'm reading from Galian, 8th ed. page 210, in case that matters). But $49 \equiv -1 \pmod{50}$, so that $\phi(-1)=6$. But now we can use the same property as before to note that $[\phi(-1)]^2 = \phi(1)=6.$ which would imply that $\phi (x) = 6x$.

But something in this line of reasoning is wrong, because $7\times 6$ is 42, which is not $6 \bmod 15$. It's $12$.

Now, I already know that the right definition for $\phi$ is that $\phi (x)= 3x$, as $3$ is the smallest integer $a$ so that $7a\equiv 6 \pmod{15}$, what I don't know and can't figure out, is why my analysis is wrong the first time.

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    $\begingroup$ Homomorphisms of groups, rings or what? $\endgroup$ – wisefool Oct 4 '14 at 23:05
  • $\begingroup$ Oh! Sorry. I thought that might be clear. Homormorphism of groups. $\endgroup$ – Alfred Yerger Oct 4 '14 at 23:06
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Because the operation in $\mathbb{Z}_{50}$ is supposed to be the sum and not the product… it is not a ring homomorphism, but just a group homomorphism and those are groups with respect to the sum! A group homomorphism preserves only an operation and in this case it is the sum, not the product!

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  • $\begingroup$ Oh I see. I have mixed up the operation in the group. Thanks for that! $\endgroup$ – Alfred Yerger Oct 4 '14 at 23:08
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$Z_{50}$ and $Z_6$ are addictive groups, so it's not true that $\phi (7)=6 \implies [\phi (7)]^2=6^2 \implies \phi(49)=6$

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