How to find $\sigma(\bar 9)$ if $\sigma(\bar5)=\bar6?$

Question

1)Let $\sigma:\Bbb Z_{12}\to\Bbb Z_{48}$ be a group homomorsiphism . find $\sigma(\bar 9)$ if $\sigma(\bar5)=\bar6$ ?

2)Let $\sigma:\Bbb Z_{10}\to\Bbb Z_{15}^*$ be a group homomorsiphism; $\bar a\in \Bbb Z_{10}$ and $\sigma(\bar a )=\bar 8$ find $\sigma(\bar {-a})$

at first question since $\sigma(\bar1)=a$ , I guess we need to solve $5a\equiv6 (mod\quad48)$ should I continue like $5a\equiv1 (mod\quad 48)$ and $48x+5y=1$ and solve this?

I have no idea about second one. can we say since it is a homo. $-\sigma(\bar a)=\sigma(\bar {-a})$

Answer 1

For the first question, your intuition is right: using your notations, $\bar 5 a = \bar 6 \pmod {48}$, so $a = (\bar 5) ^{-1} \bar 6 \pmod {48}$ ($5$ is invertible modulo $48$ because $\gcd (5, 48) = 1$). Now, $\bar 1 = \overline {48} \cdot \bar 3 + \bar 1 = \overline {145} = \bar 5 \cdot \overline {29}$, so $(\bar 5) ^{-1} = \overline {29} \pmod {48}$, hence $\sigma ( \bar 1) = a = \overline {29} \cdot \bar 6 = 30 \pmod {48}$. It follows that $\sigma (\bar 9) = \bar 9 \cdot \sigma (\bar 1) = \overline {30}$.

Concerning the second question, there are two possibilities.

1) If there is no mistyping, then, as you suspected, $\sigma (- \bar a) = \sigma (a) ^{-1} = (\bar 8) ^{-1}$ i.e. you want $x$ such that $x \cdot \bar 8 = 1 \pmod {15}$, so $x = \bar 2 \pmod {15}$.

2) If the $*$ is a mistyping, then $\sigma (- \bar a) = -\sigma (\bar a) = - \bar 8 \equiv \bar 7 \pmod {15}$.

Answer 2

For the first one, I would instead solve $$5a=9\pmod{12},$$ so that $$\sigma\left(\overline 9\right)=a\sigma\left(\overline 5\right).$$ (Do you see why?)

For the second, bear in mind that $\Bbb Z_{15}^*$--the group of units of $\Bbb Z_{15}$-- is a group under multiplication, not addition. Hence, rather than $\sigma\left(-\overline a\right)=-\sigma\left(\overline a\right),$ we can say that $$\sigma\left(-\overline a\right)=\left(\sigma\left(\overline a\right)\right)^{-1}.$$