Tour:Mind's eye test two (beginners)

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Inverses

Consider the following chain of statements. Come up with the reason in your mind, with short justifications; you should be able to write down the justifications without further thought.

1. In a monoid, any left inverse to an element must equal any right inverse.
2. In a monoid, the set of elements that have left inverses is a submonoid (with the same identity element).
3. If every element in a monoid has a left inverse, then the monoid is a group.

Using statement (3), give an alternative, more minimal definition of group.

Cancellation

This requires a little playing around with set theory, particularly the fact that for finite sets, injective maps are bijective. Solving these the first time may require some thought; however, the solutions, once obtained, are easy to store in the mind's eye.

1. A quasigroup is a magma $S$ with binary operation $*$, where for any $a,b \in S$, there exist unique $x,y \in S$ such that $a * x = y * a = b$. In a quasigroup, every element is both left and right cancellative.
2. For a finite magma (a magma whose underlying set is finite), being a quasigroup is equivalent to every element being cancellative.
3. For a semigroup, if every element is cancellative, then the semigroup is a group (split this in two parts: find an identity element, and find inverses).
4. Thus, a finite cancellative semigroup is a group.

This gives an alternative proof that any subsemigroup of a finite group, is a subgroup.

Associativity and inverses: more

These again require some thought, and possibly a review of the proof ideas we've already seen, but should be easy to hold in the mind's eye once cracked.

1. An algebra loop is a quasigroup with a neutral element (identity element) that we call $e$. Formulate two possible definitions of subloop of an algebra loop (one, postulating only closure under the binary operation, and the other, postulating that it contains the identity element). Prove that the two definitions are equivalent.
2. Suppose $(S,*)$ is an algebra loop with the property that every element $a \in S$ has a unique two-sided inverse $a^{-1} \in S$, such that we have: $a * (a^{-1} * c) = (c * a^{-1}) * a = c$ for all $c \in S$. Prove that the inverse map satisfies the reversal law, i.e. $(a * b)^{-1} = b^{-1} * a^{-1}$.

Subset-hereditariness

These are ready-to-verify statements, that require practically zero justification:

1. Associativity inherits to subsets: thus, any subset of a semigroup, closed under the multiplication, is a subsemigroup.
2. Cancellation inherits to subsets; thus, any subset of a cancellative magma, closed under the binary operation, is cancellative.
3. The existence of inverse elements doesn't inherit to subsets. So, we can find a subset of a group, closed under the multiplication, that is not closed under taking inverses.