In the solution to this problem I've been given by my professor, he says that $g^2$ generates G if n is odd but not if n is even. I am confused about this since I came up with the example that $Z/8$ is generated by 3 and also by 9, since 9 is congruent to 1, but this contradicts his answer. Any help/guidance would be much appreciated.
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2The operation of $\mathbb Z / 8\mathbb Z$ is addition modulo $8$. So "squaring" in this context corresponds to "adding twice", not literal squaring, so your $g^2$ would be $3 + 3 = 6 \mod 8$ and indeed $6$ doesn't generate the group. – balddraz May 20 '23 at 15:07
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If $g$ has order $n$, then $g^k$ generates $G$ if $\gcd(n,k)=1$ - see this duplicate for example. So if $n$ is odd, it is true for $k=2$. – Dietrich Burde May 20 '23 at 15:08
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Does this answer your question? If $g$ is the generator of a group $G$, order $n$, when is $g^k$ a generator? – Sassatelli Giulio May 20 '23 at 16:51
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You are confused because $g^2$ doesn't mean "$g$ to the power of $2$", the concept of "power" isn't even well defined for an arbitrary group. Instead, it is literally defined as $gg$. Now in $\mathbb{Z}/8\mathbb{Z}$ addition is the group operation. And so it becomes $g+g$. With that you have $3+3=6$ and $6$ is of order $4$, thus it doesn't generate $\mathbb{Z}/8\mathbb{Z}$.
freakish
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