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I'm starting to learn abstract algebra and I'm following A book of abstract algebra of C.Pinter where in chapter four he proposes the systems of equations

\begin{align} x^2=b\\ x^5=e \end{align}

where $e$ is the identity and $x$ and $b$ are elements of a group. Now it shows that we can solve for $x$ i.e. theses conditions implie $x=(b^2)^{-1}$. But the first step is

$$x^2=b\implies x^4=b^2.$$

Now here is where I have the problem, this kind of steps are not necesarily revertible. For example if we were working on $\mathbb{Z}_4$ then we could have $$4\cdot 4=2\cdot 2$$ and it would be false that $4=2$.

And the thing when we solve equations is that we arrive at a solution that satisfies, i.e. makes true the conditions; but seeing this I can't see why that value of $x$ would satisfy the conditions.

I don't know if I'm terribly misunderstading a concept or if I'm overthinking the excercise. The real question is, is this reasoning right? Does it makes sense to solve equations in groups and other structures that doesn't behave very well in this sense? Any clarification is appreciated.

J. W. Tanner
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Miguel Forteza
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    "these conditions imply $x=(b^2)^{-1}$" indeed, and "this kind of steps are not necesarily revertible" indeed, and Pinter certainly didn't claim they are. Read his complete solution more thoroughly. – Anne Bauval May 21 '25 at 21:31
  • That's exactly what $\Rightarrow$ means. Since we know $x^2=b$, we can conclude $x^4=b^2$ from it. The other direction would not be possible, but is also not needed for the proof. – Stefan May 21 '25 at 21:32
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    The first two commenters have addressed your misunderstanding, but I want to say how great it is that you're asking these questions, and being suspicious of what's said in the text -- when those suspicions are addressed, you're wiser because of it, and having puzzled over it will make the lesson more salient in the future. In short: keep questioning, keep asking about things that confuse you! – John Hughes May 21 '25 at 21:37
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    $\mathbb Z_4$ is not a group under multiplication. – Thomas Andrews May 21 '25 at 22:37
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    However, under addition, it is a group, and $3+3=1+1$ in that group. You can only reverse $x^4=b^2$ ibecause $x^5=1$ means $x^4=x^{-1}.$ You can't do it without knowing something other than $x^2=b.$ – Thomas Andrews May 21 '25 at 22:39

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(When reading the following, notice that we never take roots. We don't even "divide" -- although multiplying on the left by an inverse element is really, really close to dividing.)

It may help to see this written out a bit more. Suppose $x^2 = b$. Then \begin{align*} x^4 &= x^3 \cdot x & \text{definition of power of element} \\ &= (x^2 \cdot x) \cdot x & \text{definition of power of element} \\ &= x^2 \cdot (x \cdot x) & \text{associativity} \\ &= x^2 \cdot x^2 & \text{definition of power of element} \\ &= b \cdot x^2 & \text{properties of equality for $x^2 = b$} \\ &= b \cdot b & \text{properties of equality for $x^2 = b$} \\ &= b^2 & \text{definition of power of element} \end{align*}

For more on that property of equality, see https://math.stackexchange.com/a/3679842/123905 .

So we have shown $x^2 = b \implies x^4 = b^2$. However, you might believe that Pinter uses that inference in reverse. He does not. The line of inference is \begin{align*} e &= x^5 & \text{given} \\ &= x^4 \cdot x & \text{definition of power of element} \\ &= b^2 \cdot x & \text{properties of equality for $x^4 = b^2$} \end{align*} Now using left-multiplication (do not make the mistake of using "inside multiplication" (multiplying on the left and right at the equality) or "outside multiplication" (multiplying on the left of the left-hand side of the equation and on the right of the right-hand side of the equation), associativity, simplification of multiplication by the identity in the group, and the definition of inverses in a group, ... \begin{align*} e &= b^2 \cdot x \\ (b^2)^{-1} \cdot e &= (b^2)^{-1} \cdot (b^2 \cdot x) \\ (b^2)^{-1} &= (b^2)^{-1} \cdot (b^2 \cdot x) \\ (b^2)^{-1} &= ((b^2)^{-1} \cdot b^2) \cdot x \\ (b^2)^{-1} &= (e) \cdot x \\ (b^2)^{-1} &= x \end{align*} Then, by the symmetric property of equality, $x = (b^2)^{-1}$.

Eric Towers
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