I realize this question was asked before, but I did not find the answers satisfying. Here is my attempt:

Since G is cyclic, any element can be written as $g^m$ for $ 0 \leq m \leq 11 $, so the equation reads $ (g^k)^2 = g^j $. These two elements are equal if $ 2k \equiv j \mod 12 $ which implies $j = 2(m - 6 z)$, so if we want to find a $g$, which does not then it has to be an element with $j$ odd.

However, the solution says that $ g^6 = (g^2)^6 = e $, which is a contradiction.

  • $\begingroup$ You seek an element $g$ which is not a square. $\endgroup$
    – Wuestenfux
    Commented May 4, 2019 at 16:09
  • 2
    $\begingroup$ Why not just write down all the squares and see what elements show up? $\endgroup$ Commented May 4, 2019 at 16:10
  • $\begingroup$ @Wuestenfux Ok, so I was supposed to solve for x and not g? But it says find a g? $\endgroup$
    – ngc1300
    Commented May 4, 2019 at 16:19
  • $\begingroup$ @EthanBolker The square of every element will give me g^0, g^2, g^4, g^6, g^8, g^10. So... doesn't this agree with what I have? There can't be any odd powered elements. $\endgroup$
    – ngc1300
    Commented May 4, 2019 at 16:24
  • $\begingroup$ That group is $\{0,1,2,\ldots,11\}$ with addition modulo $12$. With that operation squaring is doubling. So double all the elements modulo $12$ and see what's missing. Part of your confusion is confusing the $g$ that's not a square with a generator of the group. $\endgroup$ Commented May 4, 2019 at 16:32

2 Answers 2


Using additive notations, you are looking for $n\in\mathbb Z_{12}$ such that for all $k\in\mathbb Z_{12}$, $n\neq 2k$. If $1$ is not such a number, then there exists $k$ such that $2k=1$, so that $2$ is invertible in the ring $\mathbb Z_{12}$, but this is not the case because $2$ and $12$ are not coprime.


Put another way, let $g$ be an element that generates $C$; as $C$ is cyclic there indeed exists such a $g$.

Note that in a cyclic group $C$, for any $d$ that divides the order $n$ of $C$, the set of $d$-th powers is a subgroup of $C$ of order $n/d$. So putting $d=2$ and $n=12$, the set of squares $S$ of $C$ is a subgroup of $C$ of order 12/2 = 6.

So suppose $g$ is in $S$. Then by the fact that $S$ is a group, every power of $g$ is also in $S$, which (as $g$ generates the whole group $C)$ inpies that $|S| = |C| = 12$. This contradicts with what we observed in the paragraph above that $|S| = 6$.

So $g$ is not in $S$, and thus there is no element that squares to $g$.

  • $\begingroup$ So then, in the problem statement, $g$ does not refer to any element of $G$, $g$ has to be a generator also? For example, $g^4$ is in $G$, but does not generate $G$. If that is not the case that $g$ also has to be a generator, then I can find a counter example to your last statement $g^4 = (g^2)^2$. $g^2$ and $g^4$ are both in $S$. $\endgroup$
    – ngc1300
    Commented May 4, 2019 at 21:56

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