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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Therefore $\displaystyle G$ 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 $\displaystyle (M, p, 1)$ was a monoid with multiplication $\displaystyle p$.

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.