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Given a semigroup $(R^1,.)$ does there exist a ring say $(R,+,.)$ such that the semigroup $(R,.)$ is same as $(R^1,.)$.

My professor said take $R^1=\{7\}$ and define $7. 7=7$.

Then the above proposition fails.There does not exist any ring $(R,+,.)$ such that the semigroup $(R,.)$ is same as $(R^1,.)$.

How is it true? I don't get his point.

Is it correct?

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    You need an absorbing element. – 2017-02-07

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Take $R=\{7\}$

Define $7\oplus 7=7$.

Now note that $(R,\oplus)$ is a group (why?)

Obviously it is closed .

The identity element is $7$ and so is the inverse element.

Note the distributive laws

$7. (7\oplus 7)=7.7\oplus 7.7 $ also hold

Hence $(R,\oplus ,.)$ is the required ring.

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    "The required ring"? The requirement was to prove there aren't any such rings :) But saying that it *is* in fact a ring is important either way. – 2017-02-07
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    I wanted to show his professor's claim is false ;isn't it clear from my answer @rschwieb – 2017-02-07
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    That intention is not totally clear, no, but I acknowledge that it is a good point, and I guessed you were trying to make it :) – 2017-02-07
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The proposed example is isomorphic to $\{0\}$ which is a ring, so I don't see the point of the professor's example either.

It's easy to find semigroups that don't have two-sided absorbing elements (but a ring always has an absorbing element in its semigroup: $0$.)

For example, take $a=a^2=ab$ and $b=b^2=ba$. Then $\{a,b\}$ is a semigroup with no two-sided absorbing element.