# Thread: Algebra, Problems For Fun (10)

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

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

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

2. by Counting by order of elements theorem since $\displaystyle n$ is a divisor of $\displaystyle n$ $\displaystyle \implies\$ number of elements in G of order n is a multiple of $\displaystyle \varphi($$\displaystyle n) \displaystyle \implies\ \displaystyle n=k*\varphi(n) for some \displaystyle k=positive integer. gcd(\displaystyle n,\varphi(n))=gcd(\displaystyle k*\varphi(n),\varphi(n))=1 for some \displaystyle k=positive integer. \displaystyle \implies\ \displaystyle \varphi($$\displaystyle n)=1$ $\displaystyle \implies\$ $\displaystyle n=1$ or $\displaystyle n=2$
if ord(G)=1 then G={e} $\displaystyle \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 $\displaystyle n$ is a divisor of $\displaystyle n$ $\displaystyle \implies\$ number of elements in G of order n is a multiple of $\displaystyle \varphi($$\displaystyle n) \displaystyle \implies\ \displaystyle n=k*\varphi(n) for some \displaystyle k=positive integer. gcd(\displaystyle n,\varphi(n))=gcd(\displaystyle k*\varphi(n),\varphi(n))=1 for some \displaystyle k=positive integer. \displaystyle \implies\ \displaystyle \varphi($$\displaystyle n)=1$ $\displaystyle \implies\$ $\displaystyle n=1$ or $\displaystyle n=2$
if ord(G)=1 then G={e} $\displaystyle \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 $\displaystyle n$ with $\displaystyle \gcd(n, \varphi(n))=1.$ for example if $\displaystyle n$ 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: ($\displaystyle \varphi(n)$ here is the Euler totient function)

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

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

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

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