Determine/guess the identity element and show that it is in fact the identity element.
Determine/guess the inverse of each element and show that it is in fact the inverse.
Proof.
The product of $f_{a_1, b_1}$ and $f_{a_2, b_2}$ is given by
\[ f_{a_2, b_2}(x)\circ f_{a_1, b_1}(x) =a_2(a_1x + b_1)+b_2=a_2a_1 x +a_2b_1 + b_2.\]
Since the group operation is function composition, it is associative.
The identity element is $f_{1,0}=x$ since for any $f_{a,b} \in G$, we have
\[f_{a,b}(x)\circ f_{1,0}(x)=af_{1,0}(x)+b=ax+b=f_{a,b}(x)\] and
\[f_{1,0}(x) \circ f_{a,b}(x)=x\circ f_{a,b}(x)=f_{a,b}(x).\]
Now we find the inverse of $f_{a_1,b_1}$.
If $f_{a_2, b_2}(x)\circ f_{a_1, b_1}(x)=x(=f_{1,0})$, then we should have $a_2 a_1=1$ and $a_2 b_1 +b_2=0$. Solving these, we obtain $a_2=1/a_{1}>0$ and $b_2=-a_2 b_1=-b_1/a_{1}$.
Thus our candidate of the inverse $f_{a,b}^{-1}$ is
\[f_{1/a_1, -b_1/a_1}=x/a_1-b_1/a_{1}.\]
In fact it is the inverse since it also satisfies
\[f_{a_1,b_1} \circ f_{1/a_1, -b_1/a_1}=a_1 (x/a_1-b_1/a_{1})+b_1 =x-b_1+b_1=x.\]
Thus $f_{a,b}^{-1}=f_{1/a_1, -b_1/a_1}$ and we conclude that $G$ is a group.
The Order of $ab$ and $ba$ in a Group are the Same
Let $G$ be a finite group. Let $a, b$ be elements of $G$.
Prove that the order of $ab$ is equal to the order of $ba$.
(Of course do not assume that $G$ is an abelian group.)
Proof.
Let $n$ and $m$ be the order of $ab$ and $ba$, respectively. That is,
\[(ab)^n=e, […]
Non-Abelian Group of Order $pq$ and its Sylow Subgroups
Let $G$ be a non-abelian group of order $pq$, where $p, q$ are prime numbers satisfying $q \equiv 1 \pmod p$.
Prove that a $q$-Sylow subgroup of $G$ is normal and the number of $p$-Sylow subgroups are $q$.
Hint.
Use Sylow's theorem. To review Sylow's theorem, check […]
Non-Abelian Simple Group is Equal to its Commutator Subgroup
Let $G$ be a non-abelian simple group. Let $D(G)=[G,G]$ be the commutator subgroup of $G$. Show that $G=D(G)$.
Definitions/Hint.
We first recall relevant definitions.
A group is called simple if its normal subgroups are either the trivial subgroup or the group […]
Order of Product of Two Elements in a Group
Let $G$ be a group. Let $a$ and $b$ be elements of $G$.
If the order of $a, b$ are $m, n$ respectively, then is it true that the order of the product $ab$ divides $mn$? If so give a proof. If not, give a counterexample.
Proof.
We claim that it is not true. As a […]
The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$
Let $p$ be a prime number.
Let $G$ be a non-abelian $p$-group.
Show that the index of the center of $G$ is divisible by $p^2$.
Proof.
Suppose the order of the group $G$ is $p^a$, for some $a \in \Z$.
Let $Z(G)$ be the center of $G$. Since $Z(G)$ is a subgroup of $G$, the order […]
Group of Invertible Matrices Over a Finite Field and its Stabilizer
Let $\F_p$ be the finite field of $p$ elements, where $p$ is a prime number.
Let $G_n=\GL_n(\F_p)$ be the group of $n\times n$ invertible matrices with entries in the field $\F_p$. As usual in linear algebra, we may regard the elements of $G_n$ as linear transformations on $\F_p^n$, […]
Isomorphism Criterion of Semidirect Product of Groups
Let $A$, $B$ be groups. Let $\phi:B \to \Aut(A)$ be a group homomorphism.
The semidirect product $A \rtimes_{\phi} B$ with respect to $\phi$ is a group whose underlying set is $A \times B$ with group operation
\[(a_1, b_1)\cdot (a_2, b_2)=(a_1\phi(b_1)(a_2), b_1b_2),\]
where $a_i […]