I have been thinking about this question for a while now.

Let $G$ be a group defined by the group of integers modulo $5$, meaning $$G=\mathbb{Z}_5=\{0,1,2,3,4\}.$$ For every $a,b$ in $G$ we define the operation:$$a\oplus b=(a+b+2)\bmod 5.$$ Find all the generators of $G$.

I know that $({\small 1})$ if $\langle H,\ast\rangle$ is a cyclic group of prime order, then every element except for the identity element, is a generator of $H$.

So for the modulo group $\mathbb{Z}_5$ under addition, the generators are $\{1,2,3,4\}$.

My questions are:

  • Now assuming that the identity element of the group $G$ is $3$, does that means that the generators of the group $G$ are $\{0,1,2,4\}$?
  • And if so, how do I verify the generators of any other cyclic group?
  • Assuming that I do not know $({\small 1})$, how could I approach the question differently?

Sorry for asking too many questions in one post. Answering the first two questions would be more than enough.

  • $\begingroup$ What are your arguments concerning the assertion that $(G,\oplus)$ is a group? $\endgroup$ – Jens Schwaiger Jun 8 '18 at 10:08
  • $\begingroup$ @JensSchwaiger It is a given statement by the question, "Let $G$ be a group...". $\endgroup$ – Bshara Zahran Jun 8 '18 at 16:15
  • $\begingroup$ The statement is that $G$ is the group of integers mod 5! But now I see the second answer. Transport of structure is, as often, the key. $\endgroup$ – Jens Schwaiger Jun 9 '18 at 3:17

Firstly, any group of prime order has to by cyclic. Indeed, any element which is not the identity generates the group. So, if the identity element of the group is $3$ then the others have to be the generators. For the second question, a cyclic group of order $n$ has exactly $\phi(n)$ generators, where $\phi$ is Euler's totient function. To find the generators, say the cyclic group is $\{a_i\}_1^n$. Now an element $a^k$ is a generator if and only if $k$ is coprime to $n$. This is not hard to show, and showing this will also tell that the number of generators is $\phi(n)$. As for the last question, you can always find the subgroup generated by each element and check if it's the full group, but that's lame. Anyways, the fact that the order of an element always divides the order of the group is very important, so you should know (1) anyways.

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The operation is $G$ is $a\oplus b= f^{-1}(f(a)+f(b))$, where $f: G \to \mathbb Z_5$ is given by $f(x)=x+2$.

Note that $f$ is a bijection. This construction is called transport of structure: it transports the group structure of $\mathbb Z_5$ to the set $G$ such that $G$ is automatically a group isomorphic to $\mathbb Z_5$.

In particular, $G$ is cyclic and its generators are $f^{-1}(u)$, where $u$ is a generator of $\mathbb Z_5$. All nonzero elements of $\mathbb Z_5$ are generators. Thus, all elements of $G$ different from $f^{-1}(0)=3$ are generators.

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  • $\begingroup$ Well, this also answers my third question, thank you! $\endgroup$ – Bshara Zahran Jun 8 '18 at 16:13

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