# Thread: Algebra, Problems For Fun (10)

1. ## Algebra, Problems For Fun (10)

Okay, this one is easy but I really like it: ( $\varphi(n)$ here is the Euler totient function)

Let $G$ be a group of order $n.$ Prove that if $\gcd(n, \varphi(n))=1,$ then $G$ is abelian.

2. by Counting by order of elements theorem since $n$ is a divisor of $n$ $\implies\$ number of elements in G of order n is a multiple of $\varphi($ $n)$ $\implies\$ $n=k*\varphi(n)$ for some $k=$positive integer.
gcd( $n,\varphi(n)$)=gcd( $k*\varphi(n),\varphi(n)$)=1 for some $k=$positive integer.
$\implies\$ $\varphi($ $n)=1$ $\implies\$ $n=1$ or $n=2$
if ord(G)=1 then G={e} $\implies\$ G is abelian
if ord(G)=2 then G is abelian since if a,b in G then a*b=b*a because (a=b) or (either a or b equal e).

is this correct?

3. Originally Posted by jaco
by Counting by order of elements theorem since $n$ is a divisor of $n$ $\implies\$ number of elements in G of order n is a multiple of $\varphi($ $n)$ $\implies\$ $n=k*\varphi(n)$ for some $k=$positive integer.
gcd( $n,\varphi(n)$)=gcd( $k*\varphi(n),\varphi(n)$)=1 for some $k=$positive integer.
$\implies\$ $\varphi($ $n)=1$ $\implies\$ $n=1$ or $n=2$
if ord(G)=1 then G={e} $\implies\$ G is abelian
if ord(G)=2 then G is abelian since if a,b in G then a*b=b*a because (a=b) or (either a or b equal e).

is this correct?
no it's not! there are infinitely many integers $n$ with $\gcd(n, \varphi(n))=1.$ for example if $n$ is any prime number. you need Sylow theorems to solve the problem.

4. Originally Posted by NonCommAlg
Okay, this one is easy but I really like it: ( $\varphi(n)$ here is the Euler totient function)

Let $G$ be a group of order $n.$ Prove that if $\gcd(n, \varphi(n))=1,$ then $G$ is abelian.
here's a solution: excluding the trivial case $n=1,$ we must have $n=p_1p_2 \cdots p_k,$ where $p_j$ are distinct primes, since $\gcd(n,\varphi(n))=1.$ in this case $\varphi(n)=(p_1-1)(p_2-1) \cdots (p_k - 1).$ now

for any $j \leq k,$ let $n_j$ be the number of Sylow $p_j$ subgroups of $G.$ suppose $n_j > 1.$ then since $n_j \mid p_i,$ for some $i \neq j,$ we must have $n_j=p_i.$ but we also have that $p_j \mid n_j - 1 = p_i - 1,$ which

is impossible becasue then $p_j \mid n$ and $p_j \mid \varphi(n).$ thus $n_j=1$ for all $j,$ i.e. every Sylow subgroup of $G$ is normal. let $P_j$ be the Sylow $p_j$ subgroup of $G.$ then:

$G \cong P_1 \times \cdots \times P_k \cong \mathbb{Z}/p_1 \mathbb{Z} \times \cdots \times \mathbb{Z}/p_k \mathbb{Z} \cong \mathbb{Z}/p_1 \cdots p_k \mathbb{Z}=\mathbb{Z}/n \mathbb{Z}.$ so we actually proved that $G$ is more than abelian, it's cyclic!