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I need to show that $(\mathbb{Z}_4 \ (0), \times)$ with multiplication modulo 5 has 2 generators; $2$ and $4$.
I understand why $2$ is a generator, but I don't get why $4$ is.

Edit: I'm not sure that's formatting correctly but it should be the set of non-zero integers with addition modulo 4.

I figured, as it is multiplication modulo five you would do
$4^1 = 4$, $4^2=1$, $4^3 = 4$, $4^4=1$
so I fail to see how this could be a generator as I was under the impression you needed to produce all the elements.

I worked out the generator $2$ in the same way so unless it's a coincindence that I got the answer correct I don't see why it doesn't work for 4.

Alice
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1 Answers1

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You are considering the group that is often denoted $\mathbb{Z}_5^\times$. Some people don't like the $\times$ because they think of rings (and you might not know about them yet). Since it is modulo $5$, you have $4$ the same as $-1$, so $4$ is not a generator for this group. $2$ is a generator because $$ 2^1 = 2 \\ 2^2 = 4 \\ 2^3 = 3 \\ 2^4 = 1. $$ So you get everything in the group. If you check, then you will find that $3$ is also a generator.

Thomas
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