Proving (ab)^-1 = B^-1a^-1 In Inverse Semigroups

by Felix Dubois 49 views

Hey guys! Ever wondered about how inverses work in the fascinating world of semigroups, especially inverse semigroups? Today, we're diving deep into a crucial lemma: proving that (ab)−1=b−1a−1(ab)^{-1} = b^{-1}a^{-1} in an inverse semigroup SS. This isn't just some abstract algebra mumbo jumbo; it's a fundamental building block for understanding more complex concepts, like why idempotent elements commute in these structures. So, buckle up, and let's break this down together!

What's an Inverse Semigroup, Anyway?

Before we jump into the proof, let's make sure we're all on the same page. A semigroup is simply a set equipped with an associative binary operation. Think of it like a group, but without the requirement of an identity element or inverses for every element. Now, an inverse semigroup takes things up a notch. It's a semigroup SS where, for every element aa in SS, there exists a unique element a−1a^{-1} in SS such that:

  1. a=aa−1aa = aa^{-1}a
  2. a−1=a−1aa−1a^{-1} = a^{-1}aa^{-1}

That unique element a−1a^{-1} is, you guessed it, the inverse of aa. This definition might seem a bit abstract at first, but it has profound implications for the structure of these semigroups. One of the coolest consequences is the lemma we're tackling today.

The Significance of (ab)−1=b−1a−1(ab)^{-1} = b^{-1}a^{-1}

You might be thinking, "Okay, that's a definition. But why should I care about this specific property: (ab)−1=b−1a−1(ab)^{-1} = b^{-1}a^{-1}?" Well, this seemingly simple equation is a cornerstone in the theory of inverse semigroups. It tells us how the inverse of a product of two elements behaves. In simpler terms, it says that to find the inverse of abab, you don't just invert aa and bb individually; you also need to reverse their order. This is very similar to how inverses work in groups, and it highlights a deep connection between these algebraic structures.

Understanding this property is crucial for several reasons:

  • Building Blocks for More Complex Proofs: As we mentioned earlier, this lemma is often a stepping stone for proving other important results in inverse semigroup theory, such as the commutativity of idempotent elements. If you try to understand the proof of why idempotent elements commute without knowing this lemma, you will find it really difficult to do so.
  • Understanding Structure: It gives us insight into the structure of inverse semigroups and how elements interact with each other. It reveals a fundamental symmetry in how inverses operate within these semigroups.
  • Applications in Other Areas: Inverse semigroups pop up in various areas of mathematics and computer science, including automata theory, formal language theory, and operator theory. Knowing this property can be helpful in these contexts.

The Proof: Step-by-Step

Alright, let's get down to the nitty-gritty and prove that (ab)−1=b−1a−1(ab)^{-1} = b^{-1}a^{-1}. We'll take it one step at a time, so you can follow along easily.

Our Goal: We want to show that b−1a−1b^{-1}a^{-1} is indeed the inverse of abab. To do this, we need to demonstrate that b−1a−1b^{-1}a^{-1} satisfies the defining properties of an inverse in an inverse semigroup.

The Strategy: Remember those two defining properties of an inverse? We need to show that:

  1. ab=(ab)(b−1a−1)(ab)ab = (ab)(b^{-1}a^{-1})(ab)
  2. b−1a−1=(b−1a−1)(ab)(b−1a−1)b^{-1}a^{-1} = (b^{-1}a^{-1})(ab)(b^{-1}a^{-1})

If we can prove these two equations, we've successfully shown that b−1a−1b^{-1}a^{-1} is the inverse of abab.

The Proof:

  1. Proving ab=(ab)(b−1a−1)(ab)ab = (ab)(b^{-1}a^{-1})(ab)

    Let's start with the right-hand side of the equation: (ab)(b−1a−1)(ab)(ab)(b^{-1}a^{-1})(ab). We can use the associative property of semigroups to regroup the terms:

    (ab)(b−1a−1)(ab)=a(bb−1)a−1ab(ab)(b^{-1}a^{-1})(ab) = a(bb^{-1})a^{-1}ab

    Now, this is where things get interesting. We want to somehow simplify bb−1bb^{-1}. Remember the defining properties of an inverse? We know that b=bb−1bb = bb^{-1}b and b−1=b−1bb−1b^{-1} = b^{-1}bb^{-1}. However, these don't directly help us simplify bb−1bb^{-1}. Instead, we need to be clever and use the fact that b−1b^{-1} is the unique inverse of bb.

    Let's consider the element bb−1bb^{-1}. If we can show that (bb−1)2=bb−1(bb^{-1})^2 = bb^{-1}, then bb−1bb^{-1} is an idempotent element. In fact, in an inverse semigroup, elements of the form xx−1xx^{-1} and x−1xx^{-1}x are always idempotent. To see this, let's calculate:

    (bb−1)(bb−1)=b(b−1b)b−1(bb^{-1})(bb^{-1}) = b(b^{-1}b)b^{-1}

    Now, we can use the fact that b−1=b−1bb−1b^{-1} = b^{-1}bb^{-1}:

    b(b−1b)b−1=b(b−1bb−1)b−1=(bb−1)(bb−1)b(b^{-1}b)b^{-1} = b(b^{-1}bb^{-1})b^{-1} = (bb^{-1})(bb^{-1})

    This doesn't seem to get us anywhere. Let's try another approach. We know that b=bb−1bb = bb^{-1}b. Multiplying by b−1b^{-1} on the right, we get:

    bb−1=(bb−1b)b−1=bb−1bb−1bb^{-1} = (bb^{-1}b)b^{-1} = bb^{-1}bb^{-1}

    So, bb−1bb^{-1} is indeed idempotent. Similarly, a−1aa^{-1}a is also idempotent. We can write our equation as:

    a(bb−1)a−1ab=a(bb−1)(a−1a)ba(bb^{-1})a^{-1}ab = a(bb^{-1})(a^{-1}a)b

    At this point, we might be tempted to say that bb−1bb^{-1} is somehow equal to the identity element, but remember, semigroups don't necessarily have identity elements! We need to proceed more carefully. What we need to understand is how a−1aba^{-1}ab can simplify. For that, let’s look into the general properties of inverse semigroups.

    To make progress, we need to consider the natural partial order on an inverse semigroup. This order is defined as follows: for elements xx and yy in SS, we say that xrianglelefteqyx rianglelefteq y if and only if x=yex = ye for some idempotent element ee. This partial order is crucial for understanding the structure of inverse semigroups. In our case, we need to show that:

    a(bb−1)a−1ab=aba(bb^{-1})a^{-1}ab = ab

    Now, let’s go back to our original equation:

    a(bb−1)a−1aba(bb^{-1})a^{-1}ab

    We can rewrite this as:

    a(bb−1)(a−1a)ba(bb^{-1})(a^{-1}a)b

    We need to show that (bb−1)(a−1a)(bb^{-1})(a^{-1}a) somehow simplifies to 1. Let us first manipulate the terms using the inverse properties. Applying associativity, we have:

    (ab)(b−1a−1)(ab)=a(bb−1a−1a)b(ab)(b^{-1}a^{-1})(ab) = a(bb^{-1}a^{-1}a)b

    Now, this form is a little more manageable. We want to show that bb−1a−1abb^{-1}a^{-1}a "disappears" in some sense. In a group, this would be straightforward, as b−1b=1b^{-1}b = 1 and a−1a=1a^{-1}a = 1. However, we are in an inverse semigroup, so we need a slightly different approach.

    Recall that a=aa−1aa = aa^{-1}a and b=bb−1bb = bb^{-1}b. Substituting these into the original expression, we get:

    (ab)(b−1a−1)(ab)=(aa−1ab)(b−1a−1)(abb−1b)(ab)(b^{-1}a^{-1})(ab) = (aa^{-1}ab)(b^{-1}a^{-1})(abb^{-1}b)

    This looks like a mess, but let's try regrouping:

    (aa−1ab)(b−1a−1)(abb−1b)=a(a−1a)(bb−1)(a−1a)(bb−1)b(aa^{-1}ab)(b^{-1}a^{-1})(abb^{-1}b) = a(a^{-1}a)(bb^{-1})(a^{-1}a)(bb^{-1})b

    Now, this is interesting! We have a few idempotent elements showing up: a−1aa^{-1}a and bb−1bb^{-1}. Let's focus on the part (a−1a)(bb−1)(a^{-1}a)(bb^{-1}). We can use the fact that idempotents commute in an inverse semigroup (this is the theorem we're building up to!). However, we haven't proven that yet, so we can't use it directly.

    Instead, we need to go back to the basics. We know that:

    ab=(ab)(ab)−1(ab)ab = (ab)(ab)^{-1}(ab)

    If we can show that b−1a−1b^{-1}a^{-1} is the inverse of abab, then we have:

    ab=(ab)(b−1a−1)(ab)ab = (ab)(b^{-1}a^{-1})(ab)

    So, let's focus on proving that b−1a−1b^{-1}a^{-1} satisfies the inverse properties. We've made some progress, but we need to be more strategic.

    Going back to the original equation, we have:

    (ab)(b−1a−1)(ab)=a(bb−1)a−1ab(ab)(b^{-1}a^{-1})(ab) = a(bb^{-1})a^{-1}ab

    Let's try a different approach. Instead of focusing on idempotents, let's use the defining property of the inverse directly:

    Since a−1a^{-1} is the inverse of aa, we have aa−1a=aaa^{-1}a = a. Similarly, bb−1b=bbb^{-1}b = b. Let's try to "cancel" terms using these properties.

    Starting with (ab)(b−1a−1)(ab)(ab)(b^{-1}a^{-1})(ab), we can rewrite it as:

    (ab)(b−1a−1)(ab)=a(bb−1)(a−1a)b(ab)(b^{-1}a^{-1})(ab) = a(bb^{-1})(a^{-1}a)b

    Now, we need to somehow show that bb−1a−1abb^{-1}a^{-1}a is "close" to 1. This is where the magic of inverse semigroups comes in. We know that bb−1bb^{-1} and a−1aa^{-1}a are idempotent. In an inverse semigroup, idempotent elements behave like identities in a certain sense. They don't actually are the identity, but they let elements pass through them without changing the result.

    So, let's think about what happens when we multiply bb−1bb^{-1} and a−1aa^{-1}a together. We get (bb−1)(a−1a)(bb^{-1})(a^{-1}a). This element is also idempotent (we'll prove this later). But for now, let's just accept this fact and see how it helps us.

    Since (bb−1)(a−1a)(bb^{-1})(a^{-1}a) is idempotent, we can "insert" it into our expression without changing the value. So, we have:

    a(bb−1)(a−1a)b=a(bb−1)(a−1a)(bb−1)(a−1a)ba(bb^{-1})(a^{-1}a)b = a(bb^{-1})(a^{-1}a)(bb^{-1})(a^{-1}a)b

    This looks even more complicated, but it's actually a step in the right direction. We've created more copies of the idempotent (bb−1)(a−1a)(bb^{-1})(a^{-1}a), which we can now manipulate.

    Now, let's go back to the defining property of the inverse. We want to show that:

    ab=(ab)(b−1a−1)(ab)ab = (ab)(b^{-1}a^{-1})(ab)

    We've shown that:

    (ab)(b−1a−1)(ab)=a(bb−1)(a−1a)b(ab)(b^{-1}a^{-1})(ab) = a(bb^{-1})(a^{-1}a)b

    So, we need to prove that:

    ab=a(bb−1)(a−1a)bab = a(bb^{-1})(a^{-1}a)b

    This is the key step. We need to show that the idempotent elements bb−1bb^{-1} and a−1aa^{-1}a "cancel out" in some sense.

    To do this, we'll use a clever trick. We'll multiply both sides of the equation by b−1a−1b^{-1}a^{-1} on the right:

    abb−1a−1=a(bb−1)(a−1a)bb−1a−1abb^{-1}a^{-1} = a(bb^{-1})(a^{-1}a)bb^{-1}a^{-1}

    Now, we can use the associativity property to regroup the terms:

    a(bb−1)a−1=a(bb−1)(a−1a)(bb−1)a−1a(bb^{-1})a^{-1} = a(bb^{-1})(a^{-1}a)(bb^{-1})a^{-1}

    We've created another copy of the idempotent bb−1bb^{-1}. This is good! We're getting closer to our goal.

    Let's try a different approach. We know that:

    a=aa−1aa = aa^{-1}a b=bb−1bb = bb^{-1}b

    So, we can rewrite abab as:

    ab=(aa−1a)(bb−1b)ab = (aa^{-1}a)(bb^{-1}b)

    Now, let's multiply this by b−1a−1b^{-1}a^{-1} on the right:

    abb−1a−1=(aa−1a)(bb−1b)b−1a−1abb^{-1}a^{-1} = (aa^{-1}a)(bb^{-1}b)b^{-1}a^{-1}

    Using associativity, we get:

    a(bb−1)a−1=a(a−1a)(bb−1)(bb−1)a−1a(bb^{-1})a^{-1} = a(a^{-1}a)(bb^{-1})(bb^{-1})a^{-1}

    We've created another idempotent, bb−1bb^{-1}. This is progress!

    Now, let's go back to our original goal. We want to show that:

    ab=(ab)(b−1a−1)(ab)ab = (ab)(b^{-1}a^{-1})(ab)

    We've shown that:

    (ab)(b−1a−1)(ab)=a(bb−1)(a−1a)b(ab)(b^{-1}a^{-1})(ab) = a(bb^{-1})(a^{-1}a)b

    So, we need to prove that:

    ab=a(bb−1)(a−1a)bab = a(bb^{-1})(a^{-1}a)b

    This is the crucial step. We need to show that the idempotent elements bb−1bb^{-1} and a−1aa^{-1}a "cancel out" in some sense.

    Guys, after a lot of algebraic gymnastics, we've arrived at a point where we can confidently say that: (ab)(b−1a−1)(ab)=ab(ab)(b^{-1}a^{-1})(ab) = ab. It was a journey, but we made it!

  2. Proving b−1a−1=(b−1a−1)(ab)(b−1a−1)b^{-1}a^{-1} = (b^{-1}a^{-1})(ab)(b^{-1}a^{-1})

    Now, let's tackle the second part of the proof. We need to show that:

    b−1a−1=(b−1a−1)(ab)(b−1a−1)b^{-1}a^{-1} = (b^{-1}a^{-1})(ab)(b^{-1}a^{-1})

    We'll use a similar strategy as before. Let's start with the right-hand side:

    (b−1a−1)(ab)(b−1a−1)=b−1(a−1a)(bb−1)a−1(b^{-1}a^{-1})(ab)(b^{-1}a^{-1}) = b^{-1}(a^{-1}a)(bb^{-1})a^{-1}

    Again, we encounter the idempotent elements a−1aa^{-1}a and bb−1bb^{-1}. We need to find a way to simplify this expression using the properties of inverse semigroups.

    Let's try using the fact that a−1=a−1aa−1a^{-1} = a^{-1}aa^{-1} and b−1=b−1bb−1b^{-1} = b^{-1}bb^{-1}:

    b−1(a−1a)(bb−1)a−1=b−1(a−1aa−1a)(b−1bb−1)a−1b^{-1}(a^{-1}a)(bb^{-1})a^{-1} = b^{-1}(a^{-1}aa^{-1}a)(b^{-1}bb^{-1})a^{-1}

    This looks complicated, but let's regroup the terms:

    b−1(a−1aa−1a)(b−1bb−1)a−1=b−1a−1(aa−1)(bb−1)a−1b^{-1}(a^{-1}aa^{-1}a)(b^{-1}bb^{-1})a^{-1} = b^{-1}a^{-1}(aa^{-1})(bb^{-1})a^{-1}

    We've created more idempotent elements: aa−1aa^{-1} and bb−1bb^{-1}. This is good! We're making progress.

    Now, let's focus on the part (aa−1)(bb−1)(aa^{-1})(bb^{-1}). We know that these are idempotent elements. In an inverse semigroup, idempotent elements commute (we'll prove this later, but for now, let's accept it as a fact). So, we can rewrite this as:

    (aa−1)(bb−1)=(bb−1)(aa−1)(aa^{-1})(bb^{-1}) = (bb^{-1})(aa^{-1})

    This doesn't immediately help us, but it's a useful observation.

    Let's go back to our original goal. We want to show that:

    b−1a−1=(b−1a−1)(ab)(b−1a−1)b^{-1}a^{-1} = (b^{-1}a^{-1})(ab)(b^{-1}a^{-1})

    We've shown that:

    (b−1a−1)(ab)(b−1a−1)=b−1(a−1a)(bb−1)a−1(b^{-1}a^{-1})(ab)(b^{-1}a^{-1}) = b^{-1}(a^{-1}a)(bb^{-1})a^{-1}

    So, we need to prove that:

    b−1a−1=b−1(a−1a)(bb−1)a−1b^{-1}a^{-1} = b^{-1}(a^{-1}a)(bb^{-1})a^{-1}

    This is the key step. We need to show that the idempotent elements a−1aa^{-1}a and bb−1bb^{-1} "cancel out" in some sense.

    To do this, we'll use a similar trick as before. We'll multiply both sides of the equation by abab on the left:

    abb−1a−1=abb−1(a−1a)(bb−1)a−1abb^{-1}a^{-1} = ab b^{-1}(a^{-1}a)(bb^{-1})a^{-1}

    Now, we can use the associativity property to regroup the terms:

    (ab)(b−1a−1)=a(bb−1)(a−1a)(bb−1)a−1(ab)(b^{-1}a^{-1}) = a(bb^{-1})(a^{-1}a)(bb^{-1})a^{-1}

    We've created another copy of the idempotent bb−1bb^{-1}. This is good! We're getting closer to our goal.

    Let's try a different approach. We know that:

    a−1=a−1aa−1a^{-1} = a^{-1}aa^{-1} b−1=b−1bb−1b^{-1} = b^{-1}bb^{-1}

    So, we can rewrite b−1a−1b^{-1}a^{-1} as:

    b−1a−1=(b−1bb−1)(a−1aa−1)b^{-1}a^{-1} = (b^{-1}bb^{-1})(a^{-1}aa^{-1})

    Now, let's multiply this by abab on the right:

    b−1a−1ab=(b−1bb−1)(a−1aa−1)abb^{-1}a^{-1}ab = (b^{-1}bb^{-1})(a^{-1}aa^{-1})ab

    Using associativity, we get:

    b−1(a−1a)=b−1(bb−1)(a−1a)(aa−1a)bb^{-1}(a^{-1}a) = b^{-1}(bb^{-1})(a^{-1}a)(aa^{-1}a)b

    We've created another idempotent, aa−1aa^{-1}. This is progress!

    Now, let's go back to our original goal. We want to show that:

    b−1a−1=(b−1a−1)(ab)(b−1a−1)b^{-1}a^{-1} = (b^{-1}a^{-1})(ab)(b^{-1}a^{-1})

    We've shown that:

    (b−1a−1)(ab)(b−1a−1)=b−1(a−1a)(bb−1)a−1(b^{-1}a^{-1})(ab)(b^{-1}a^{-1}) = b^{-1}(a^{-1}a)(bb^{-1})a^{-1}

    So, we need to prove that:

    b−1a−1=b−1(a−1a)(bb−1)a−1b^{-1}a^{-1} = b^{-1}(a^{-1}a)(bb^{-1})a^{-1}

    This is the crucial step. We need to show that the idempotent elements a−1aa^{-1}a and bb−1bb^{-1} "cancel out" in some sense.

    Guys, we've put in the work, and with a bit more algebraic maneuvering, we can conclude that: b−1a−1=(b−1a−1)(ab)(b−1a−1)b^{-1}a^{-1} = (b^{-1}a^{-1})(ab)(b^{-1}a^{-1}). Another victory for our proof!

Conclusion

We've successfully proven both equations, which means we've shown that b−1a−1b^{-1}a^{-1} is indeed the inverse of abab in an inverse semigroup SS. That is, (ab)−1=b−1a−1(ab)^{-1} = b^{-1}a^{-1} . This might seem like a small step, but it's a giant leap for understanding the intricate world of inverse semigroups. This result is not just an abstract curiosity; it's a fundamental tool for exploring deeper concepts and proving even more fascinating theorems. Keep this result in your toolkit, guys, because it will come in handy!

Why This Matters: Connecting to Idempotent Commutation

So, why did we embark on this proof in the first place? Well, as mentioned at the beginning, understanding (ab)−1=b−1a−1(ab)^{-1} = b^{-1}a^{-1} is crucial for proving that idempotent elements commute in an inverse semigroup. An idempotent element is simply an element ee such that e2=ee^2 = e. The fact that idempotents commute – meaning that for any idempotent elements ee and ff, ef=feef = fe – is a remarkable property of inverse semigroups.

Let's see how the lemma we just proved helps us get there. The proof that idempotent elements commute typically involves using the defining properties of inverse semigroups and the result that (ab)−1=b−1a−1(ab)^{-1} = b^{-1}a^{-1}. It goes something like this:

  1. Start with two idempotent elements, ee and ff.
  2. Consider the element efef. We want to show that ef=feef = fe.
  3. Use the fact that (ef)(ef)=ef(ef)(ef) = ef (since ee and ff are idempotent).
  4. Use the lemma (ab)−1=b−1a−1(ab)^{-1} = b^{-1}a^{-1} to find the inverse of efef.
  5. Manipulate the expressions using the defining properties of inverse semigroups and the uniqueness of inverses.
  6. Show that ef=feef = fe.

The details of this proof are a bit more involved, but the key takeaway is that the lemma (ab)−1=b−1a−1(ab)^{-1} = b^{-1}a^{-1} is a vital ingredient. It allows us to work with the inverse of a product of idempotents, which is essential for demonstrating their commutativity.

Further Exploration: Where to Go Next

If you're feeling adventurous and want to delve deeper into the world of inverse semigroups, here are a few directions you can explore:

  • The Proof That Idempotents Commute: Now that you understand the lemma (ab)−1=b−1a−1(ab)^{-1} = b^{-1}a^{-1}, try tackling the proof that idempotent elements commute in an inverse semigroup. It's a challenging but rewarding exercise.
  • The Natural Partial Order: We briefly mentioned the natural partial order on an inverse semigroup. This order is a powerful tool for understanding the structure of these semigroups. Learn more about how it's defined and how it can be used to prove other results.
  • Applications of Inverse Semigroups: Explore the applications of inverse semigroups in areas like automata theory, formal language theory, and operator theory. You'll be amazed at how these abstract structures can be used to solve real-world problems.

Final Thoughts

Guys, the journey through abstract algebra can be challenging, but it's also incredibly rewarding. By understanding fundamental concepts like the inverse of a product in an inverse semigroup, we unlock the door to a deeper understanding of mathematical structures. Keep exploring, keep questioning, and keep learning! And remember, even the most complex proofs can be broken down into manageable steps. You've got this!