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 $1 \leq d < c \leq 48,$

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


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

\begin{align} \phi_a (\bar{c}) & = \phi_a (\bar{d}) \\ \Longrightarrow x^{a(c - d)} & = 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 $c - d < 48$. So we must have that $c = d$.

  • $\begingroup$ $48 \mid a(c-d)$. Since $(a,48)=1 \Rightarrow 48 \mid (c - d) \Rightarrow c-d=0 \Rightarrow c=d$ which means it's injective $\endgroup$
    – Merdanov
    Commented Jun 10, 2017 at 11:25
  • $\begingroup$ @Merdanov Oh, I see the mistake I made towards the end. I have one concern, though. Isn't it also true that $(48, c - d) = 1$, since $1 \leq d < c \leq 48$ implies that $c - d < 48$. I essentially used this observation, rather than $(48, a) = 1$ as derived above, to get the apparent contradiction. $\endgroup$ Commented Jun 10, 2017 at 11:39
  • $\begingroup$ yeah if you say $d\lt c$ you get an apparent contradiction. However, the convention for showing an injection is that if $f(x)=f(y) \Rightarrow x=y$ $\endgroup$
    – Merdanov
    Commented Jun 10, 2017 at 11:42
  • $\begingroup$ @Merdanov True. Messed up towards the end. Thanks. $\endgroup$ Commented Jun 10, 2017 at 11:43
  • $\begingroup$ @Merdanov Just realized that by assuming $1 \leq d < c \leq 48$ and $\phi_a(\bar{c})$ = $\phi_a(\bar{d})$, I was implicitly giving a proof of injectivity by contradiction, not directly. That lead to the error in reasoning. $\endgroup$ Commented Jun 10, 2017 at 11:47

1 Answer 1


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 $a, n$.

  • $\begingroup$ Isn't it true that if $n \mid ab$ and $(n,b) = 1$, then $n \mid a$. I have used this theorem. $\endgroup$ Commented Jun 10, 2017 at 11:37
  • $\begingroup$ That statement is true, but I don't see how you used it: You wrote "Since $48 \nmid (c - d)$, we must have that $48 \mid a$" but did not previously show that $(n, c - d) = 1$. $\endgroup$ Commented Jun 10, 2017 at 11:47
  • $\begingroup$ Yes, I used it wrongly. See my comments above below the question. I had assumed that $d < c \leq 48$, as stated, so I concluded that $48 \nmid c -d$. $\endgroup$ Commented Jun 10, 2017 at 11:49
  • $\begingroup$ It's still not quite clear to me what you mean. You certainly can take $1 \leq d < c < 48$ without loss of generality, in which case it will be true that $48 \nmid (c - d)$, but this does not guarantee that $(48, c - d) = 1$. Consider, for example, $c = 24, d = 0$. $\endgroup$ Commented Jun 10, 2017 at 11:53
  • $\begingroup$ Sorry, I realized my mistake. I have edited my post for what I believe is a correct proof of injectivity, based on the comment in the question. $\endgroup$ Commented Jun 10, 2017 at 11:59

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