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In our Abstract Algebra class, we have just discussed the idea of a unit in a general ring and has posed the following question: "Prove that there cannot be a commutative ring with $1$ (the multiplicative identity) and $5$ units".

My assumption was that there are no commutative rings that have an odd amount of units. This is because if $a$ is a unit, then there is a $b \in R$ such that $ab = ba = 1$. But, wouldn't this imply that $b$ would have to be a unit as well? And also, I know that if $a$ is a unit, then $-a$ is a unit as well. So, this would imply that units have to come in pairs and thus, we can never have an odd amount of units. However, this person has constructed a ring with an odd amount of units.

So, I guess my question is why can I not have $5$ units in a commutative ring? What about the commutativity gives me the fact that I cannot have $5$ units?

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    You stopped writing too soon! Play around with your idea some more! – 2017-02-19
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    I have played around with this a bit, but I just don't know where to go or even if I am done. Let's say $a$ was a unit, and therefore $-a$ is a unit. This would also imply that $ab = ba = 1$ and $(-a)(-b) = ab = 1$. So, I have found 4 units: $a, b, -a,$ and $-b$. Then, I would need just one more. So, since they come in pairs, is it not possible and am I done? – 2017-02-19
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    @baconator you're forgetting about $1$ which is a unit. As proven in that link, if $R$ has an odd number of units then $1=-1$. Now you need to reason why there isn't a field with $5$ units in which $a=-a$ – 2017-02-19
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    @baconator But why do you assume that $\,a,b,-a,-b\,$ are *distinct*? – 2017-02-19
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    @BillDubuque Right, I cannot assume that they are distinct because in a ring, I could have that $a a = 1$. – 2017-02-19

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Your link actually gives a huge hint as to how to do this, as it shows that $a=-a$ in any counterexample, and more generally that $U(R)$ has no subgroups of order $2$.

Assume that $R$ is a communitive ring with unity and $5$ units. Our five units are $1, a, a^{-1}, b, b^{-1}$ for some $a,b$. To see that $a,a^{-1}$ are distinct, notice that, for all $x\in R$,

$$a=a^{-1}\Rightarrow xa=xa^{-1} \Rightarrow xa^{2}=x\Rightarrow a^2=1$$

Thus if $a\neq 1$, then $\{1,a\}\leq U(R)$ has order $2$, which is a contradiction. To see that $a\neq b, a\neq b^{-1}$ just note that if this wasn't to hold, then we just did a bad job of picking $b$, and there are two other units out there by assumption.

Now you need to show that no matter how multiplication is defined between the units, you can always derive a contradiction. Try looking at the possible values of $ab$ and then constructing an element whose square is $1$.

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    Thank you for the help! I have a question about your response, because I do not see how you can claim that $\{ 1, a \} \leq U(R)$ from the line above it. When you have that $xa^2 = x$, how can you claim that it is a subgroup? I understand why it can't be a subgroup, but why from the above line can you claim it is a subgroup? – 2017-02-19
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    @baconator that chain equality holds for all $x$, so it shows that $a^2$ is the multiplicative identity. Since the multiplicative identity is unique, $a^2=1$, making $\{1,a\}$ a subgroup of order two. It's addition and multiplication tables look exactly like that of the group ${1,-1\}$, with $a$ playing the role of $-1$. – 2017-02-19
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    @baconator I've edited the post to hopefully make this more clear – 2017-02-19
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    Because it would be cyclic, got it. Thank you! – 2017-02-19