Cyclic groups question Show that $\mathbb Z_{35}^\times$ is not cyclic.
I assume that I need to show that no element of $\mathbb Z_{35}$ has a particular order, indicating it is not cyclic, but I'm not sure how to do this.
 A: Note that $\overline{6}$ and $\overline{29}$ generate two distinct subgroups of order $2$. In a cyclic group this cannot happen.
A: Here's a generalization of the original post. We will show that if $m=ab$ for coprime $a$ and $b$, where $a$ and $b$ are odd positive integers ($a,b\neq 1$), then the group $\mathbf{Z}_{m}^{\times}$ is not cyclic.  Note that it suffices to show that if there exists $k$ such that $u^k\equiv 1\pmod{a}$, then $u^k\equiv 1\pmod{b}$, for $u\in\mathbf{Z}_{ab}^{\times}$ and $k<\varphi(m)$. We will show the result for odd primes $a$ and $b$. Since $m=ab$ we have $\varphi(m)=\varphi(ab)=\varphi(a)\varphi(b)=(a-1)(b-1)$. By Fermat's Little Theorem, $u^{a-1}\equiv 1\pmod{a}$ and $u^{b-1}\equiv 1\pmod{b}$, where $u\in\mathbf{Z}_{m}^{\times}$. Thus, we need to find $k<(a-1)(b-1)$ such that $u^k\equiv 1\pmod{ab}$. If we take $k=\text{lcm}[a-1,b-1]$, then due to the fact that $$\text{lcm}[a-1,b-1]=\frac{(a-1)(b-1)}{\gcd(a-1,b-1)},$$ we have $\text{lcm}[a-1,b-1]< (a-1)(b-1)$. We have $\text{ord}_{m}(u)|(a-1)$ and $\text{ord}_{m}(u)|(b-1)$, so $\text{ord}_{m}(u)|\text{lcm}[a-1,b-1]$, so we have $$u^{\text{lcm}[a-1,b-1]}\equiv 1\pmod{ab},$$ which is what we wanted.
It follows that $\mathbf{Z}_{35}^{\times}=\mathbf{Z}_{7\times5}^{\times}$ is not cyclic.
A: Use the Chinese remainder theorem that $\def\Z{\Bbb Z}\Z/35\Z\cong(\Z/5\Z)\times(\Z/7\Z)$ as rings since $5$ and $7$ are coprime; this implies for their multiplicative groups that $(\Z/35\Z)^\times\cong(\Z/5\Z)^\times\times(\Z/7\Z)^\times$. Now you should know that $(\Z/5\Z)^\times$ and $(\Z/7\Z)^\times$ are cyclic of orders $4$, $6$ respectively, and you can deduce that $x^{12}=1$ for any $x\in(\Z/35\Z)^\times$.
A: Besides your thoughts on the problem and another post, you may use the following facts:
$$\mathbb Z_{35}^*\cong\mathbb Z_5^*\times\mathbb Z_7^*\cong\mathbb Z_4\times\mathbb Z_6$$ Now the problem is change to this one stating if the rightmost group has an element of order $24$ or not.
A: Your group $G=\mathbb Z_{35}^{\times}$ has order $24$, right? So all you have to do is show that none of the $24$ elements of $G$ has order $24$. Let's just do it.
Let's start by picking an element of $G$ at random and computing its order. I picked $2$, and computed: $2^1=2$, $2^2=4$, $2^3=8$, $2^4=16$, $2^5$=32, $2^6=29$, $2^7=23$, $2^8=11$, $2^9=22$, $2^{10}=9$, $2^{11}=18$, $2^{12}=1$, so $2$ has order $12$.
Now we are half done! Each of those $12$ powers of $2$ has order $12$ or less, seeing as $(2^n)^{12}=(2^{12})^n=1^n=1$. Let$$H=\{2^n:n\in\mathbb Z\}=\{2^n:1\le k\le12\}=\{1,\ 2,\ 4,\ 8,\ 9.\ 11,\ 16,\ 18,\ 22,\ 23,\ 29,\ 32\}.$$We know that $H$ is a subgroup of $G$ (right?) and $h^{12}=1$ for every $h\in H$.
Now let's pick a random element of $G\setminus H$. I picked $3$. What's the order of $3$? Well, $3^2=9=2^{10}$, so $3^{12}=2^{60}=1^5=1$, so the order of $3$ is $12$. Now the coset $$3H=\{3h:h\in H\}$=\{3,\ 6,\ 12,\ 24,\ 27,\ 33,\ 13,\ 19,\ 31,\ 34,\ 17,\ 31\}=\{3,\ 6,\ 12,\ 13,\ 17,\ 19,\ 24,\ 26,\ 27,\ 31,\ 33,\ 34\}$$contains the other $12$ elements of $G$. If $x\in3H$, then $x^{12}=(3h)^{12}=3^{12}h^{12}=1\cdot1=1.$
We have shown that every element of $G$ satisfies the equation $x^{12}=1$. That means that every element has order $1$, $2$, $3$, $4$, $6$, or $12$. There are no elements of order $24$, so $G$ is cyclic!
