# Thread: 2 Basic algebra questions

1. ## 2 Basic algebra questions

I am studying by my self Basic Algebra by Nathan Jacobson, and I am currently doing the exercises from chapter 1.
I would be very thankful if someone would look over my answers and
tell me if my argument is sound.

1. Let $(M,p,1)$ be a monoid with multiplication p, and let $m\in M$. Define a new product $p_m$ in $M$ by $p_m(a,b)=amb$.
Show that this defines a semigroup. Under what conditions on $m$ do we have a unit relative to $p_m$?

For this to be a semigroup then we have to show that $p_m$ is associative. Let $*$ denote $p_m$.
Let $a,b,c \in M$, then

$(a*b)*c=(amb)mc=am(bmc)=a*(b*c)$, because composition under $p$ is associative, so $p_m$ is associative so this is a semigroup.

Now we do have a unit relative to $p_m$ if and only if $m$ has an inverse under $p$. Then the unit relative to $p_m$ is $m^{-1}$.
That is

$a*m^{-1}=amm^{-1}=a1=a$ and $m^{-1}*a=m^{-1}ma=1a=a$ for all $a\in M$

2. Show that $G$ cannot be the union of two proper subgroups.

Lets assume that $G$ is the union of proper subgroups $A,B$. Since they are proper there exist elements $a\in A$, $b\in B$ such that $a\notin B$ and $b\notin A$.
Now since $G$ is a group then it is closed under composition, that is $ab\in G$. Then $ab\in A$ or $ab\in B$.

Since $A$ is a subgroup and $a\in A$ then $a^{-1} \in A$ and $A$ is closed under composition. Then $a^{-1}\cdot ab=b\in A$ but that is a contradiction. In the same way
$ab\in B$ leads to $a\in B$ which is also a contradiction.

Therefore $G$ cannot be the union of two proper subgroups.

Thank you.
Hjörtur

2. Originally Posted by hjortur
I am studying by my self Basic Algebra by Nathan Jacobson, and I am currently doing the exercises from chapter 1.
I would be very thankful if someone would look over my answers and
tell me if my argument is sound.

1. Let $(M,p,1)$ and let $m\in M$. Define a new product $p_m$ in $M$ by $p_m(a,b)=amb$.
Show that this defines a semigroup. Under what conditions on $m$ do we have a unit relative to $p_m$?

For this to be a semigroup then we have to show that $p_m$ is associative. Let $*$ denote $p_m$.
Let $a,b,c \in M$, then

$(a*b)*c=(amb)mc=am(bmc)=a*(b*c)$, because composition under $p$ is associative, so $p_m$ is associative so this is a semigroup.

Now we do have a unit relative to $p_m$ if and only if $m$ has an inverse under $p$. Then the unit relative to $p_m$ is $m^{-1}$.
That is

$a*m^{-1}=amm^{-1}=a1=a$ and $m^{-1}*a=m^{-1}ma=1a=a$ for all $a\in M$

2. Show that $G$ cannot be the union of two proper subgroups.

Lets assume that $G$ is the union of proper subgroups $A,B$. Since they are proper there exist elements $a\in A$, $b\in B$ such that $a\notin B$ and $b\notin A$.
Now since $G$ is a group then it is closed under composition, that is $ab\in G$. Then $ab\in A$ or $ab\in B$.

Since $A$ is a subgroup and $a\in A$ then $a^{-1} \in A$ and $A$ is closed under composition. Then $a^{-1}\cdot ab=b\in A$ but that is a contradiction. In the same way
$ab\in B$ leads to $a\in B$ which is also a contradiction.

Therefore $G$ cannot be the union of two proper subgroups.

Thank you.
Hjörtur
That seems good to me, although it would have helped if you mentioned that $(M, p, 1)$ was a monoid with multiplication $p$.

3. Thank you, I edited the post to fix this. I dont know how I could forget to type the latter half of the sentence.