Let $G$ be an abelian group.
Let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively.
Prove that there exists an element $c$ in $G$ such that the order of $c$ is the least common multiple of $m$ and $n$.

Also determine whether the statement is true if $G$ is a non-abelian group.

Hint.

First, consider the case when $m$ and $n$ are relatively prime.

Proof.

When $m$ and $n$ are relatively prime

Recall that if the orders $m, n$ of elements $a, b$ of an abelian group are relatively prime, then the order of the product $ab$ is $mn$.
(For a proof, see the post “Order of the Product of Two Elements in an Abelian Group“.)

So if $m, n$ are relatively prime, then we can take $c=ab\in G$ and $c$ has order $mn$, which is the least common multiple.

The general Case

Now we consider the general case.

Let $p_i$ be the prime factors of either $m$ or $n$.
Then write prime factorizations of $m$ and $n$ as
\[m=\prod_{i}p_i^{\alpha_i} \text{ and } n=\prod_{i} p_i^{\beta_i}.\] Here $\alpha_i$ and $\beta_i$ are nonzero integers (could be zero).
Define
\[m’=\prod_{i: \alpha_i \geq \beta_i}p_i^{\alpha_i} \text{ and } n’=\prod_{i: \beta_i> \alpha_i} p_i^{\beta_i}.\]

(For example, if $m=2^3\cdot 3^2\cdot 5$ and $n=3^2\cdot 7$ then $m’=2^3\cdot 3^2\cdot 5$ and $n’=7$.)

Note that $m’\mid m$ and $n’\mid n$, and also $m’$ and $n’$ are relatively prime. The least common multiple $l$ of $m$ and $n$ is given by
\[l=m’n’\]

Consider the element $a’:=a^{m/m’}$. We claim that the order of $a’$ is $m’$.

Let $k$ be the order of the element $a’$. Then we have
\begin{align*}
e=(a’)^k=(a^{\frac{m}{m’}})^k=a^{mk/m’},
\end{align*}
where $e$ is the identity element in the group $G$.
This yields that $m$ divides $mk/m’$ since $m$ is the order of $a$.
It follows that $m’$ divides $k$.

On the other hand, we have
\begin{align*}
(a^{m/m’})^{m’}=a^m=e,
\end{align*}
and hence $k$ divides $m’$ since $k$ is the order of the element $a^{m/m’}$.

As a result, we have $k=m’$.
So the order of $a’$ is $m’$.

Similarly, the order of $b’:=b^{n/n’}$ is $n’$.

The orders of elements $a’$ and $b$ are $m’$ and $n’$, and they are relatively prime.
Hence we can apply the first case and we conclude that the element $a’b’$ has order
\[m’n’=l.\] Thus, we can take $c=a’b’$.

The Case When $G$ is a Non-Abelian Group

Next, we show that if $G$ is a non-abelian group then the statement does not hold.

For example, consider the symmetric group $S_3$ with three letters.
Let
\[a=(1\,2\,3) \text{ and } b=(1 \,2).\]

Then the order of $a$ is $3$ and the order of $b$ is $2$.
The least common multiple of $2$ and $3$ is $6$.
However, the symmetric group $S_3$ have no elements of order $6$.

Hence the statement of the problem does not hold for non-abelian groups.

Related Question.

For a solution of this problem, see the post “Order of Product of Two Elements in a Group“.

The post The Existence of an Element in an Abelian Group of Order the Least Common Multiple of Two Elements first appeared on Problems in Mathematics.

Prove that every finite group having more than two elements has a nontrivial automorphism.

(Michigan State University, Abstract Algebra Qualifying Exam)
 

Proof.

Let $G$ be a finite group and $|G|> 2$.

Case When $G$ is a Non-Abelian Group

Let us first consider the case when $G$ is a non-abelian group.
Then there exist elements $g, h\in G$ such that $gh\neq hg$.

Consider the map $\phi: G \to G$ defined by sending $x\in G$ to $gxg^{-1}$.
Then it is straightforward to check that $\phi$ is a group homomorphism and its inverse is given by the conjugation by $g^{-1}$.
Hence $\phi$ is an automorphism.

If $\phi=1$, then we have $h=\phi(h)=ghg^{-1}$, and this implies that $gh=hg$.
This contradicts our choice of $g$ and $h$.
Hence $\phi$ is a non-trivial automorphism of $G$.

Case When $G$ is an Abelian Group

Next consider the case when $G$ is a finite abelian group of order greater than $2$.
Since $G$ is an abelian group the map $\psi:G\to G$ given by $x \mapsto x^{-1}$ is an isomorphism, hence an automorphism.

If $\psi$ is a trivial automorphism, then we have $x=\psi(x)=x^{-1}$.
Thus, $x^2=e$, where $e$ is the identity element of $G$.

Sub-Case When $G$ has an Element of Order $\geq 3$.

Therefore, if $G$ has at least one element of order greater than $2$, then $\psi$ is a non-trivial automorphism.

Sub-Case When Elements of $G$ has order $\leq 2$.

It remains to consider the case when $G$ is a finite abelian group such that $x^2=e$ for all elements $x\in G$.
In this case, the group $G$ is isomorphic to
\[\Zmod{2}\times \Zmod{2}\times \cdots \Zmod{2}=(\Zmod{2})^n.\] Since $|G| > 2$, we have $n>1$.

Then the map $(\Zmod{2})^n\to (\Zmod{2})^n$ defined by exchanging the first two entries
\[(x_1, x_2, x_3, \dots, x_n) \mapsto (x_2, x_1, x_3, \dots, x_n)\] is an example of nontrivial automorphism of $G$.

Therefore, in any case, the group $G$ has a nontrivial automorphism.

The post Every Finite Group Having More than Two Elements Has a Nontrivial Automorphism first appeared on Problems in Mathematics.

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 counterexample, consider $G=S_3$, the symmetric group of three letters.
Let $a=(1\, 2), b=(1 \,3)$ be transposition elements in $S_3$.
The orders of $a$ and $b$ are both $2$.

Consider the product
\[ab=(1\, 2)(1 \,3)=(1 \, 3 \, 2).\] Then it is straightforward to check that the order of $ab$ is $3$, which does not divide $4$ (the product of orders of $a$ and $b$).

Therefore, the group $G=S_3$ and elements $a=(1\, 2), b=(1 \,3)\in G$ serve as a counterexample.

Remark. (Abelian group case)

If we further assume that $G$ is an abelian group, then the statement is true.
Here is the proof if $G$ is abelian.

Let $e$ be the identity element of $G$.
\begin{align*}
(ab)^{mn} &=a^{mn}b^{mn} && \text{ since $G$ is abelian}\\
&=(a^m)^n(b^n)^m\\
&=e^n e^m && \text{since the order of $a, b$ are $m, n$ respectively}\\
&=e.
\end{align*}
Thus the order of $ab$ divides $mn$.

Related Question.

If the group is abelian, then the statement is true.

See the post “Order of the Product of Two Elements in an Abelian Group” for a proof of this problem.

More generally, we can prove the following.

A proof of this problem is given in the post “The Existence of an Element in an Abelian Group of Order the Least Common Multiple of Two Elements“.

The post Order of Product of Two Elements in a Group first appeared on Problems in Mathematics.

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 out the post Sylow’s Theorem (summary). Read the corollary there as well to understand the proof below.

Proof.

Let $n_p, n_q$ be the number of $p$-Sylow subgroups and $q$-Sylow subgroups, respectively.
Then by Sylow’s theorem, we have
\[n_p\equiv 1 \pmod p, \qquad n_p|q \tag{*}\] and
\[n_q \equiv 1 \pmod q, \qquad n_q|p. \tag{**}\] Since $q \equiv 1 \pmod p$, we have $q>p$. Thus $n_q$ must be $1$ from (**).
Hence, $G$ has a unique $q$-Sylow subgroup, and it is normal.

From (*), the possibilities for $n_p$ are either $1$ or $q$.
We eliminate the possibility of $n_p=1$ as follows.
If $n_p=1$, then $G$ has a unique $p$-Sylow subgroup, and hence it is normal.
Let $P, Q$ be the unique normal $p$-Sylow subgroup and $q$-Sylow subgroup of $G$, respectively. Then $P, Q$ are normal subgroup of order $p$ and $q$, and hence $P \cap Q=\{e\}$, where $e$ is the identity element of $G$.
Since the order $|PQ|=pq=|G|$, these conditions imply that we have
\[G\cong P \times Q.\] Since $P$ and $Q$ are groups of prime order, hence it is cyclic, in particular abelian.
The direct product of abelian group is abelian, hence $G$ is abelian.
This is a contradiction.

Thus, we must have $n_p=q$.

The post Non-Abelian Group of Order $pq$ and its Sylow Subgroups first appeared on Problems in Mathematics.

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,
\[(ab)^n=e, (ba)^m=e,\] where $e$ is the identity element of $G$.

We compute
\begin{align*}
e&=(ab)^n=\underbrace{(ab)\cdot (ab) \cdot (ab) \cdots (ab)}_{n\text{ times}}\\[6pt] &=a\underbrace{\cdot (ba)(ba)\cdot (ba) \cdots (ba)}_{n-1\text{ times}}b\\[6pt] &=a(ba)^{n-1}b.
\end{align*}

From this, we obtain
\[(ba)^{n-1}=a^{-1}b^{-1}=(ba)^{-1},\] and thus we have
\[(ba)^n=e.\] Therefore the order $m$ of $ba$ divides $n$.

Similarly, we see that $n$ divides $m$, and hence $m=n$.
Thus the orders of $ab$ and $ba$ are the same.

The post The Order of $ab$ and $ba$ in a Group are the Same first appeared on Problems in Mathematics.

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 itself.
• The commutator subgroup $D(G)=[G,G]$ is a subgroup of $G$ generated by all commutators $[a,b]=a^{-1}b^{-1}ab$ for $a,b\in G$.

The commutator subgroup $D(G)=[G,G]$ is a normal subgroup of $G$.
For a proof, see: A condition that a commutator group is a normal subgroup.

Proof.

Note that the commutator subgroup $D(G)$ is a normal subgroup.
Since $G$ is simple, any normal subgroup of $G$ is either the trivial group $\{e\}$ or $G$ itself. Thus we have either $D(G)=\{e\}$ or $D(G)=G$.
If $D(G)=\{e\}$, then for any two elements $a,b \in G$ the commutator $[a,b]\in D(G)=\{e\}$.

Thus we have
\[a^{-1}b^{-1}ab=[a,b]=e.\] Therefore we have $ab=ba$ for any $a,b\in G$. This means that the group $G$ is abelian, which contradicts with the assumption that $G$ is non-abelian.
Therefore, we must have $D(G)=G$ as required.

The post Non-Abelian Simple Group is Equal to its Commutator Subgroup first appeared on Problems in Mathematics.

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 of the center is also a power of $p$, that is, $|Z(G)|=p^b$, for some $b \in \Z$.
Then we have the index $[G: Z(G)]=p^{a-b}$.

If $a-b=0$, then we have $G=Z(G)$ and $G$ is an abelian group. This contradicts with the assumption that $G$ is non-abelian. So $a-b \neq 0$.

If $a-b=1$, then the order of the quotient $|G/Z(G)|=[G:Z(G)]=p$ is a prime, thus $G/Z(G)$ is a cyclic group.
Recall that if the quotient by the center is cyclic, then the group is abelian.
Thus the group $G$ is abelian, which again a contradiction.

Therefore, we must have $a-b \geq 2$, hence $p^2$ divides the index $[G: Z(G)]=p^{a-b}$.
This concludes the proof.

The post The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$ first appeared on Problems in Mathematics.

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 $a_i \in A, b_i \in B$ for $i=1, 2$.

Let $f: A \to A’$ and $g:B \to B’$ be group isomorphisms. Define $\phi’: B’\to \Aut(A’)$ by sending $b’ \in B’$ to $f\circ \phi(g^{-1}(b’))\circ f^{-1}$.

\[\require{AMScd}
\begin{CD}
B @>{\phi}>> \Aut(A)\\
@A{g^{-1}}AA @VV{\sigma_f}V \\
B’ @>{\phi’}>> \Aut(A’)
\end{CD}\] Here $\sigma_f:\Aut(A) \to \Aut(A’)$ is defined by $ \alpha \in \Aut(A) \mapsto f\alpha f^{-1}\in \Aut(A’)$.
Then show that
\[A \rtimes_{\phi} B \cong A’ \rtimes_{\phi’} B’.\]

Proof.

Define $\Psi: A \rtimes_{\phi} B \to A’ \rtimes_{\phi’} B’$ by
\[(a,b) \mapsto (f(a), g(b))\] for $(a,b)\in A \rtimes_{\phi} B$.
We show that this is a group isomorphism. Since $f, g$ are group isomorphisms, it suffices to show that $\Psi$ is a group homomorphism.

Let $(a_1,b_1), (a_2, b_2) \in A \rtimes_{\phi} B$. We compute the product in $A\rtimes_{\phi} B$ is
\[(a_1,b_1)\cdot(a_2, b_2)=(a_1\phi(b_1)(a_2), b_1 b_2).\] Thus we have
\begin{align*}
\Psi\left( (a_1,b_1)\cdot(a_2, b_2) \right) &= \Psi\left( (a_1\phi(b_1)(a_2), b_1 b_2) \right)\\
&=\bigg(f \big(a_1\phi(b_1)(a_2) \big), g(b_1 b_2) \bigg). \tag{*}
\end{align*}

On the other hand, we have
\begin{align*}
\Psi \left( (a_1, b_1) \right) \cdot \Psi \left( (a_2, b_2) \right) &=
(f(a_1), g(b_1)) \cdot (f(a_2), g(b_2)).\\
&=\bigg(f(a_1) \phi’\big(g(b_1))(f(a_2) \big), g(b_1)g(b_2) \bigg) \tag{**}
\end{align*}
Here we used group operation in $A’\rtimes_{\phi’} B’$ in the second equality.

Now by the definition of $\phi’$ we have
\[\phi'(g(b_1))=f\circ \phi(g^{-1}(g(b_1))) \circ f^{-1}=f\circ \phi(b_1) \circ f^{-1}.\] Thus we have
\[\phi'(g(b_1))(f(a_2))=f\circ \phi(b_1) a_2.\] Hence we have
\begin{align*}
(**) &= \bigg(f(a_1)f \big(\phi(b_1)a_2 \big), g(b_1)g(b_2) \bigg)\\
&=\bigg(f \big(a_1\phi(b_1)(a_2) \big), g(b_1 b_2)\bigg),
\end{align*}
where the last equality follows since $f, g$ are group homomorphisms.

Comparing this with (*), we see that
\[\Psi\left( (a_1,b_1)\cdot(a_2, b_2) \right)=\Psi \left( (a_1, b_1) \right) \cdot \Psi \left( (a_2, b_2) \right) \] and thus $\Psi$ is a group homomorphism, hence it is a group isomorphism.

Corollary

In particular, taking $A=A’$ and $B=B’$, we have the following corollary.

Let $A \rtimes_{\phi} B$ be the semidirect product of groups $A$ and $B$ with respect to a homomorphism $\phi: B \to \Aut(A)$.
If $\phi’ :B \to \Aut(A)$ is defined by the following diagram, then we have
\[A \rtimes_{\phi} \cong A \rtimes_{\phi’} B.\]

\[\require{AMScd}
\begin{CD}
B @>{\phi}>> \Aut(A)\\
@A{g^{-1}}AA @VV{\sigma_f}V \\
B @>{\phi’}>> \Aut(A)
\end{CD}\]

Application

Determine all isomorphism classes of semidirect product groups $(C_2 \times C_2) \rtimes C_3$, where $C_i$ denotes a cyclic group of order $i$.

Proof.

We first determine all homomorphism $\phi: C_3 \to \Aut(C_2 \times C_2)$.
Note that $\Aut(C_2 \times C_2)\cong S_3$. ($C_2 \times C_2$ has three degree $2$ elements and an automorphism permutes these elements.)

Let $g$ be a generator of $C_3$. Then since $g$ is of order $3$, the image $\phi(g)$ is one of $1$, $(123)$, $(132)$.
Thus there are three homomorphism from $C_3 \to \Aut(C_2 \times C_2)$ defined by
\[\phi_0(g)=1, \phi_1(g)=(123), \phi_2(g)=(132).\]

Since $\phi_0$ is a trivial homomorphism, the semidirect product is actually a direct product. Thus
\[(C_2\times C_2)\rtimes_{\phi_0} C_3 =(C_2\times C_2)\times C_3,\] which is an abelian group.

For $\phi_1$ and $\phi_2$, we obtain nonabelian groups
\[(C_2\times C_2)\rtimes_{\phi_1} C_3 \text{ and } (C_2\times C_2)\rtimes_{\phi_2} C_3 .\] We claim that these two groups are isomorphic.

Note that we have
\[\phi_1(x)=\tau \phi_2 (x) \tau^{-1}\] for all $x \in C_3$, where $\tau=(23)\in S_3$.
Thus the claim follows from Corollary.
(In the notations in Corollary, $A=C_2 \times C_2$, $B=C_3$, $f\in \Aut(A)\cong S_3$ is $(23)$ and $g:B\to B$ is the identity.)
Therefore we have two isomorphism classes for $(C_2 \times C_2) \rtimes C_3$, one is abelian, the other is nonabelian.
(In fact, the nonabelian group is isomorphic to $A_4$.)

The post Isomorphism Criterion of Semidirect Product of Groups first appeared on Problems in Mathematics.

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 $\F_p^n$, the $n$-dimensional vector space over $\F_p$. Therefore, $G_n$ acts on $\F_p^n$.

Let $e_n \in \F_p^n$ be the vector $(1,0, \dots,0)$.
(The so-called first standard basis vector in $\F_p^n$.)

Find the size of the $G_n$-orbit of $e_n$, and show that $\Stab_{G_n}(e_n)$ has order $|G_{n-1}|\cdot p^{n-1}$.

Conclude by induction that
\[|G_n|=p^{n^2}\prod_{i=1}^{n} \left(1-\frac{1}{p^i} \right).\]

Proof.

Let $\calO$ be the orbit of $e_n$ in $\F_p^n$.
We claim that $\calO=\F_p^n \setminus \{0\}$, hence
\[|\calO|=p^n-1.\]

To prove the claim, let $a_1 \in \F_p^n$ be a nonzero vector.
Then we can extend this vector to a basis of $\F_p^n$, that is, there is $a_2, \dots, a_n \in \F_p^n$ such that $a_1,\ a_2, \dots, a_n$ is a basis of $\F_P^n$.
Since they are a basis the matrix $A=[a_1 \dots a_n]$ is invertible, that is , $A \in G_n$.
We have
\[Ae_n=a_1\] Thus $a_1\in \calO$. It is clear that $0 \not \in \calO$. Thus we proved the claim.

Next we show that
\[|\Stab_{G_n}(e_n)|=|G_{n-1}|\cdot p^{n-1}. \tag{*} \] Note that $A \in \Stab_{G_n}(e_n)$ if and only if $A e_n=e_n$.
Thus $A$ is of the form
\[ \left[\begin{array}{r|r}
1 & A_2 \\ \hline
\mathbf{0} & A_1
\end{array} \right], \] where $A_1$ is an $(n-1)\times (n-1)$ matrix, $A_2$ is a $1\times (n-1)$ matrix , and $\mathbf{0}$ is the $(n-1) \times 1$ zero matrix.
Since $A$ is invertible, the matrix $A_1$ must be invertible as well, hence $A_1 \in G_{n-1}$.
The matrix $A_2$ can be anything.
Thus there are $|G_{n-1}|$ choices for $A_1$ and $p^{n-1}$ choices for $A_2$.
In total, there are $|G_{n-1}|p^{n-1}$ possible choices for $A \in \Stab_{G_n}(e_n)$. This proves (*).

Finally we prove that
\[|G_n|=p^{n^2}\prod_{i=1}^{n} \left(1-\frac{1}{p^i} \right)\] by induction on $n$.

When $n=1$, we have
\[|G_1|=|\F_p\setminus \{0\}|=p-1=p\left(1-\frac{1}{p} \right).\]

Now we assume that the formula is true for $n-1$.
By the orbit-stabilizer theorem, we have
\[ |G_n: \Stab_{G_n}(e_n)|=|\calO|.\] Since $G_n$ is finite, we have
\begin{align*}
|G_n|&=|\Stab_{G_n}(e_n)||\calO|\\
&=(p^n-1)|G_{n-1}|p^{n-1}\\
&=(p^n-1)p^{n-1}\cdot p^{(n-1)^2}\prod_{i=1}^{n-1} \left(1-\frac{1}{p^i} \right) \text{ by the induction hypothesis}\\
&=p^n\left(1-\frac{1}{p^n} \right)p^{n-1}p^{n^2-2n+1} \prod_{i=1}^{n-1} \left(1-\frac{1}{p^i} \right) \\
&=p^{n^2}\prod_{i=1}^{n} \left(1-\frac{1}{p^i} \right).
\end{align*}
Thus the formula is true for $n$ as well.

By induction, the formula is true for any $n$.

The post Group of Invertible Matrices Over a Finite Field and its Stabilizer first appeared on Problems in Mathematics.

Define the functions $f_{a,b}(x)=ax+b$, where $a, b \in \R$ and $a>0$.

Show that $G:=\{ f_{a,b} \mid a, b \in \R, a>0\}$ is a group . The group operation is function composition.

Steps.

Check one by one the followings.

1. The group operation on $G$ is associative.
2. Determine/guess the identity element and show that it is in fact the identity element.
3. 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.

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