Going through a book and to my satisfaction I came up with a proof for the statement...

*If a^k = e where k is an odd integer, then the order of a is odd*

Thinking of the contrapositive for this statement, how should I view the word "where"? Should I take "where" to mean "and"? If I do think of "where" as "and", then I have the conditional statement...

*If the order of a is even, then [a^k is not equal to e or k is an even integer]*

but this doesn't seem right to me. How should I negate a statement containing the word "where"?

While I'm at it, what's the correct way for negating a statement with "when" in it. Since these two words can be used interchangeably (sometimes).

The original statement would be better phrased as:

Let $k$ be an odd integer. Then $a^k = e \implies |a|$ is odd.

This makes it clear that the contrapositive is:

$|a|$ even, and $a^k = e \implies k$ is even.

In other words, "where" in the original statement actually means "only if", applied to the entire rest of the sentence, and our original statement is of the form:

$p \to (q \to r)$

where:

$p$ = $k$ is an odd integer.

$q$ = $a^k = e$.

$r$ = $|a|$ is odd.

The contrapositive of this is:

$\sim(q \to r) \to \sim p$

and it may seem strange, but the negation of an implication:

$q \to r$

is an "AND" statement:

$q\ \wedge \sim r$

One can see this by comparing truth-tables, or by using DeMorgan's laws to the disjunctive normal form of implication:

$\sim q \vee r$

or one can simply see it from this natural language example:

the negation of the statement:

"If I hit my hand with a hammer, it will hurt" is:

"I hit my hand with a hammer, and it didn't hurt".

It can be difficult to express a natural language statement "faithfully" in classical logic, there are nuances to how we express ideas in natural languages that complicate this extraction, sentential clauses can be put "out of order", as in this example (the statement about $k$, logically should come FIRST, since everything else depends on it-we cannot even evaluate $a^k$ without knowing what $k$ is).

In general, "when/where" usually indicates a dependency, which you should mentally translate as an implication.

For example:

"We know that $a^ma^n = a^{m+n}$ when(where) $n$ is a natural number"

should be translated as:

"If $n$ is a natural number, then $a^ma^n = a^{m+n}$".

Or, perhaps more pithily:

[A when(where) B] = [A if B] = [if B, then A].

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On a more general note, the point of the original statement you were reading about, is that if $|a| = m$ and $a^k = e$ for non-negative integers $k,m$, then $m|k$.

To see this, note that we cannot have $0 < k < m$, by definition of order. If $k = 0$, trivially we have $a^0 = e$, but EVERY positive integer divides 0.

So, without loss of generality assume $m \leq k$.

We can thus write:

$k = qm + r$, where $0 \leq r < m$.

If $ r \neq 0$, this leads to:

$e = a^k = a^{qm + r} = (a^{qm})(a^r) = (a^{mq})(a^r) = (a^m)^q(a^r) = e^qa^r = ea^r = a^r$, contradicting that the order of $a$ is $m$.

Thus $r = 0$ is the only possibility, so that $k = qm$ that is $m|k$.

So if $k$ is odd, then $a^k = e$ implies that the order of $a$ must be odd, because odd numbers do not have ANY factors of 2, and thus ANY even numbers, as divisors.

On the other hand, if the order of $a$ is even, then that even number divides any $k$ for which $a^k = e$, which means in particular that:

$|a| = m = 2t$ and $a^k = e$, which as we saw above means $k = mq$ (for some $q$) and this leads to $k = (2t)q = 2(tq)$, which is an even number.