primitive residue classes modulo 32 $\mathbb Z_{32}^*$ is the primitive residue classes modulo 32. How is it possible to show that $\mathbb Z_{32}^*$ is generated by 5 and -1, without showing it for every element of $\mathbb Z_{32}^*$=$\{1,3,5,7,9,11,...,31\}$
 A: The number $32$ is small, so it is easy to show by computing that $5$ has order $8$. Note that we only need to compute $5^{2^k}$ modulo $32$. For $\varphi(32)=16$, so the order of $5$ must be a power of $2$. 
The following is severe overkill. We show that for any $n\ge 3$, $\mathbb{Z}^\ast_{2^n}$ is generated by $5$ and $-1$.
Lemma: Let $n\ge 3$. Then $5^{2^{n-3}}\not\equiv 1 \pmod{2^n}$, while $5^{2^{n-2}}\equiv 1\pmod{2^n}$.
Proof of Lemma: We show by induction on $n$ that if $n \ge 3$, then $5^{2^{n-3}}\equiv 1+2^{n-1}\pmod{2^n}$. This easily yields both parts of our assertion.
The result is easy to check for $n=3$.   Suppose now that 
$$5^{2^{k-3}}\equiv 1+2^{k-1}\pmod{2^{k}}.\tag{$1$}$$ We show that $5^{2^{k-2}}\equiv 1+2^{k}\pmod{2^{k+1}}$. 
 Square both sides of $(1)$, and simplify modulo $2^{k+1}$. We get
 $$5^{2^{k-2}}=(1+2^{k-1} +n2^k)^2\equiv 1+2^{k}\pmod{2^{k+1}},$$
 since $2^{2k-2}$ is divisible by $2^{k+1}$ if $k\ge 3$.  This ends the proof of the lemma.
Now it is easy to finish.  The order of $5$ modulo $2^n$ divides $\varphi(2^n)=2^{n-1}$, so the order of $5$ is a power of $2$. It follows from the results above that the order of $5$ is exactly $2^{n-2}$.
No power of $5$ is congruent to $-1$ modulo $2^n$. It follows that if $n \ge 3$, then $5$ and $-1$ together generate all of $\mathbb{Z}^\ast_{2^n}$. 
A: Let $n > 1$ be an integer.
We consider the group $G = (\mathbb{Z}/2^n\mathbb{Z})^*$.
Clearly $|G| = 2^{n-1}$.
Let $e \ge 2$ be an integer.
Let $k$ be an odd integer.
$(1 + 2^e k)^2 = 1 + 2^{e+1}k + 2^{2e} k^2 = 1 + 2^{e+1}(k + 2^{e-1}k^2)$.
Note that $k + 2^{e-1}k^2$ is odd.
Hence, by induction on $m \ge 1$, $(1 + 2^e k)^{2^m} = 1 + 2^{e+m}k'$ for some odd integer $k'$.
Since $5 = 1 + 2^2$, $5^{2^m} = 1 + 2^{m + 2}k'$, where $k'$ is odd.
Hence the order of $5$ in $G$ is $2^{n-2}$.
Let $H$ be the subgroup of $G$ generated by $5$.
Suppose $-1 \in H$.
There exists integer $m$ such that $-1 \equiv 5^m$ (mod $2^n$).
Since $5 \equiv 1$ mod $4$, $-1 \equiv 1$ (mod $4$).
This is a contradiction.
Hence $G$ is generated by $-1$ and $5$.
A: Working modulo $\,32\,$:
$$5^2=-7\;\;,\;\;5^3=-35=-3\;\;,\;5^6=(5^3)^2=9\;\;,\;5^8=5^6\cdot 5^2=9\cdot (-7)=-63=1$$
$$\Longrightarrow 5^8=1\Longrightarrow ord_{32}(5)=8$$
$$(-1)^2=1\Longrightarrow ord_{32}(-1)=2$$
Finally, since $\,\langle 5\rangle\cap\langle -1\rangle=\{1\}\,$ , we get that
$$\langle 5\rangle\langle-1\rangle\cong\langle 5\rangle\otimes\langle -1\rangle\cong\Bbb Z_{32}^*$$
since $\left|\Bbb Z_{32}^*\right|=\phi(32)=16\,$
