Dummit and Foote, Abstract Algebra, Question 7, Cyclic Groups I'm trying to solve the following question Dummit and Foote's Abstract Algebra.


  Let $Z_{48} = \langle x \rangle$. For which integers a does the map $\phi_a : \bar{1} \mapsto x^a$ extend to an ismorphism from $\mathbb{Z}/48 \mathbb{Z}$ onto $Z_{48}$


Proof: Assume that $\phi_a$ is an isomorphism. We then must have that $| \bar{1} | = | x^a |$. Since $| \bar{1} | = 48$, then $| x^a| = 48$. Using the formula $|x^a| = 48/(48,a)$, we then must have that $(48, a) = 1$. That is, $a$ is the set of all integers such that $2,3 \nmid a$.

I think this is a sufficient condition to put on $a$. For if $2, 3 \mid a$, then $\phi_a$ will not be an isomorphism, as given by the use of the previous isomorphism theorem.

But I get a contradiction (I suppose) when I try and individually impose the conditions of homomorphism and injectivity (bijectivity) to derive a condition on $a$.

$\mathbf{\phi_a}$ is a homomorphism

We can then derive the following recrusion relation.

\begin{align}
\phi_a(\overline{n-1} \; * \; \bar{1}) = \phi_a(\bar{n}) = \phi_a(\overline{n-1})\phi_a(\bar{1}) = \phi_a(\overline{n-1})x^a    
\end{align}

Using this, we can show that $\phi(\bar{n}) = x^{na}$. So, a possible map is given by this equation. It is easily shown that if we start from this map, then it is a homomorophism.

$\mathbf{\phi_a}$ is a injective

To prove that $\phi_a$ is a bijection, it suffices to check when is $\phi_a$ an injection. We have, with $1 \leq d &lt; c \leq 48,$

\begin{align}
\phi_a (\bar{c}) &amp;  = \phi_a (\bar{d}) \\
\Longrightarrow  x^{ca} &amp; = x^{da} \\
\Longrightarrow  x^{a(c - d)} &amp; = 1 \\
\end{align}

This implies that $48 \mid a(c-d)$. Since $48 \nmid (c - d)$, we must have that $48 \mid a$. But this is in contradiction with the conditions derived above. If $a = 48$, then $2, 3 \mid 48$. Also, under the mapping $\bar{1} \mapsto x^{48}$, where $x^{48}$ is the identity in $Z_{48}$. Since a homormophism maps the idenintity of one group to the identity of the group this, this is again a contradiction.

Where am I going wrong?



Edit:

Here's a correct proof of injectivity. with $1 \leq d &lt; c \leq 48,$

\begin{align}
\phi_a (\bar{c}) &amp;  = \phi_a (\bar{d}) \\
\Longrightarrow  x^{a(c - d)} &amp; = 1 \\
\end{align}

This implies that $48 \mid a(c-d)$. Since $(48, a) = 1$, we must have that $48 \mid c - d$. But this is a contradiction, sine $c - d &lt; 48$. So we must have that $c = d$.
It's not true for general $r$ that $r \mid st$ implies $r \mid s$ or $r \mid t$ for all $s, t$; this is only true when $r$ is prime.

At any rate, to check injectivity of a group homomorphism, it's sufficient to show that its kernel is trivial. So, it's enough to show that $\phi_a(x^n) = \bar 0$ (that is, $an \equiv 0 \pmod {48}$) implies $x^n = e$, (that is, that $n \equiv 0 \pmod {48}$). But this condition on $a$ is essentially by definition coprimality of $a, n$.
It's not true for general $r$ that $r \mid st$ implies $r \mid s$ or $r \mid t$ for all $s, t$; this is only true when $r$ is prime.

At any rate, to check injectivity of a group homomorphism, it's sufficient to show that its kernel is trivial. So, it's enough to show that $\phi_a(x^n) = \bar 0$ (that is, $an \equiv 0 \pmod {48}$) implies $x^n = e$, (that is, that $n \equiv 0 \pmod {48}$). But this condition on $a$ is essentially by definition coprimality of $a, n$.