# 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.

- In a monoid, any left inverse to an element must equal any right inverse.
- In a monoid, the set of elements that have left inverses is a submonoid (with the same identity element).
- 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.

- A quasigroup is a magma with binary operation , where for any , there exist unique such that . In a quasigroup, every element is both left and right cancellative.
- For a finite magma (a magma whose underlying set is finite), being a quasigroup is equivalent to every element being cancellative.
- 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).
- 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.

- An algebra loop is a quasigroup with a neutral element (identity element) that we call . 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. - Suppose is an algebra loop with the property that every element has a unique two-sided inverse , such that we have: for all . Prove that the inverse map satisfies the reversal law, i.e. .

## Subset-hereditariness

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

- Associativity inherits to subsets: thus, any subset of a semigroup, closed under the multiplication, is a subsemigroup.
- Cancellation inherits to subsets; thus, any subset of a cancellative magma, closed under the binary operation, is cancellative.
- 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.