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.

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!
