Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 2 down vote accepted

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 $(M,p,1)$.

share|cite|improve this answer
Thank you for answering. – Adelaide Dokras May 20 '12 at 4:30
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.