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## Main Question or Discussion Point

Let ##G## be a set equipped with a binary associative operation ##\cdot##.

In both of the following situations, we have a group:

1) ##G## is not empty, and for all ##a,b\in G##, there exists an ##x,y\in G## such that ##bx=a## and ##yb=a##.

2) There exists a special element ##e\in G## such that ##xe = x## for all ##x##. And (if we fix this ##e##), then for each ##x\in G##, there exists some ##x^\prime \in G## such that ##xx^\prime = e##.

For reference, ##G## equipped with a binary associative operation is a group if there exists an element ##e\in G## such that ##xe=ex=x##. And for any ##x\in G##, there is an ##x^\prime\in G## such that ##xx^\prime = x^\prime x = e##.

I've been thinking for hours and I really can't seem to figure out why either (1) or (2) forms a group.

It is easy to show that if we have a group, then ##(1)## and ##(2)## hold.

Furthermore, I know that if ##(1)## holds, then ##(2)## holds as well. Indeed, take ##a## a special element, then we can find an ##e## such that ##ae=a## (for this special ##a##). Now, take ##b## arbitrary, then there is an ##x## such that ##b = xa = xae = be##. So we have found the right element ##e##. The existence of ##x^\prime## for every ##x## is now obvious.

So it suffices to show that ##(2)## implies that we have a group. But I really can't figure it out.

In both of the following situations, we have a group:

1) ##G## is not empty, and for all ##a,b\in G##, there exists an ##x,y\in G## such that ##bx=a## and ##yb=a##.

2) There exists a special element ##e\in G## such that ##xe = x## for all ##x##. And (if we fix this ##e##), then for each ##x\in G##, there exists some ##x^\prime \in G## such that ##xx^\prime = e##.

For reference, ##G## equipped with a binary associative operation is a group if there exists an element ##e\in G## such that ##xe=ex=x##. And for any ##x\in G##, there is an ##x^\prime\in G## such that ##xx^\prime = x^\prime x = e##.

I've been thinking for hours and I really can't seem to figure out why either (1) or (2) forms a group.

It is easy to show that if we have a group, then ##(1)## and ##(2)## hold.

Furthermore, I know that if ##(1)## holds, then ##(2)## holds as well. Indeed, take ##a## a special element, then we can find an ##e## such that ##ae=a## (for this special ##a##). Now, take ##b## arbitrary, then there is an ##x## such that ##b = xa = xae = be##. So we have found the right element ##e##. The existence of ##x^\prime## for every ##x## is now obvious.

So it suffices to show that ##(2)## implies that we have a group. But I really can't figure it out.