# Take G to be the cyclic group with 12 elements. Find an element g in G such that the equation x^2 = g has no solution

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.

• You seek an element $g$ which is not a square. Commented May 4, 2019 at 16:09
• Why not just write down all the squares and see what elements show up? Commented May 4, 2019 at 16:10
• @Wuestenfux Ok, so I was supposed to solve for x and not g? But it says find a g? Commented May 4, 2019 at 16:19
• @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. Commented May 4, 2019 at 16:24
• 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. Commented May 4, 2019 at 16:32

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 $$g$$.
• 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 $S$. Commented May 4, 2019 at 21:56