# What condition on $m$ turns this semigroup into a monoid?

Suppose you have a monoid $(M,p,1)$ (viewing it as a triple of a set $M$, operation $p$, and unit $1$). Then for some $m\in M$ we can define a new product $p_m$ in $M$ by $p_m(a,b)=amb$. It's easy to see this is a semigroup.

However, what condition on $m$ will we have a unit relative to $p_m$? If a unit $e$ were to exist, then I suppose $p_m(a,e)=p_m(e,a)=a$, that is, $ame=ema=a$.

What condition on $m$ am I supposed to be getting at? At first I thought we would require that $m$ commute with all of $M$, but the last equality above is giving me a problem. Thanks.

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It appears that $m$ needs to have a two-sided inverse in $M$. – Brian M. Scott May 20 '12 at 4:06
@BrianM.Scott Oh of course, and this two sided inverse $m^{-1}$ under $p$ is the identity under $p_m$. Thanks. – Adelaide Dokras May 20 '12 at 4:08
@Adeal: So you see why if $m^{-1}$ exists, then $(M,p_m,m^{-1})$ is a monoid. Do you also see why if $(M,p_m,e)$ is a monoid, then necessarily $m^{-1}$ exists and $e=m^{-1}$? – Jonas Meyer May 20 '12 at 4:15
@JonasMeyer That's a good point to include, thanks. I assume it follows by looking at $p_m(1,e)=p_m(e,1)=1$, so $1me=em1=1$, or $me=em=1$. – Adelaide Dokras May 20 '12 at 4:29

If $m$ is invertible in $M$, then $m^{-1}$ is an identity for $p_m$: $p_m(a,m^{-1}) = amm^{-1} = a1 = a$ and $p_m(m^{-1},a) = m^{-1}ma = 1a = a$ for all $a\in M$.
Conversely, suppose that $e$ is an identity for $p_m$. Then in particular, $1=p_m(1,e) = 1me = me$ and $1=p_m(e,1) = em1 = em$, so $em=me = 1$, hence $e$ is an inverse for $m$ in $M$.
So $(M,p_m)$ is a monoid if and only if $m$ is invertible in $(M,p,1)$.
It seems worth noting that when $(M,p)$ is not a monoid, then $(M,p_m)$ cannot be a monoid either, because if it were, then $me$ would be a right identity in $(M,p)$ and $em$ would be a left identity, and it implies that $em=me$ would be an identity in $(M,p).$ – user23211 May 20 '12 at 9:06