I have seen the prove of general theorem if $G$ is non abelian group of order $pq, p>q$ then $G$ has q linear irreducible characters. The proof uses this property: The degree of irreducible character divides the group's order. I wanna see if we set $q=2$ can we prove the theory without using the property?

"A non abelian group $G$ of order $2p$ has 2 linear irreducible characters".

Here is my attempt:

Let $G$ be a non-abelian group of order $2p$, I want to show that $[G:[G,G]]=2$ $(|[G,G]|=p)$

Since $[G,G]$ is subgroup of $G$ then by lagrange thm $|[G,G]|$ and $[G:[G,G]]$ divides $2p$

$|[G,G]| \in$ {1,2,p,2p}.

Case 1: $|[G,G]|=1 \Rightarrow G$ is abelian (contradiction)

Case 2: $|[G,G]|=2p \Rightarrow [G,G]=G$ Groups of order $pq$ are solvable where $p$ & $q$ are distinct primes and for solvable group the commutator is proper subgroup, then $|[G,G]|=2p$ is a contradiction

Case 3: $|[G,G]|=2$ Then $G$ has $p$ linear irreducible charactes. Let $k$ be the number of the irreducible characters, then $p < k <2p$. $Irr(G)=$ {$\chi_1, …, \chi_k$}

$|G| = ∑_{χ∈Irr(G)}χ(1)2 = p + ∑_{χ∈Irr(G),χ(1)≠1}χ(1)2 =2p$

$∑_{χ∈Irr(G),χ(1)≠1}χ(1)2 =p$

Since the degree of the characters is always positive integer then the minimum degree of the non- linear irreducible characters is 2, then $\chi^2(1)= ,4,9,16….$.

The number of the non-linear irreducible characters is between $1$ & $p-1$ denote them by $m$ ($1≤m≤p-1$)

Clearly $m≠1$ since primes are not squares.

If $m=p-1$:

$4≤\chi^2(1)$ for every irreducible non-linear character $\Rightarrow 4(p-1)≤∑_{χ∈Irr(G)χ(1)≠1}χ(1)2 =p \Rightarrow 3p≤4$ and this yields to contradiction.

Then $1<m<p-1$ *

Now let’s take $m=p-2, p-3, … $ where $p-i>1 \Rightarrow p>i+1$

If $m=p-2$ in the same way $4(p-2)≤p$ this yields to contradiction ($p=2>3$)

If $m=p-3$ in the same way $4(p-3)≤p$ this yields to contradiction ($p=2$ or $3 > 4$)

If $m=p-4$ in the same way $4(p-4)≤p$ this yields to contradiction($ p=2$ or $3$ or $5 >5$)

I want to show that for every integer $i > 4$, for $m=p-i$ we get contradiction

Suppose $m=p-i$ where $i>4$ like above, $4(p-i)≤∑_{χ∈Irr(G),χ(1)≠1}χ(1)2 =p \Rightarrow 4p-4i≤p \Rightarrow p≤4/3i$

I can't reach to a contradiction!!


By the Cauchy lemma, $G$ has an element $x$ of order $p$. Let $H$ be the subgroup generated by $x$, then $[G:H]=2$ so that $H$ is normal and $G/H$ abelian, so that $[G,G] \subset H$. But $H$ has prime order and $[G,G]$ is nontrivial, so that $[G,G]=H$. Therefore $G^{ab}$ has order $2$, thus $\mathrm{Hom}(G,\mathbb{C}^{\times})$ has two elements.

  • $\begingroup$ Can't we do the same for 2. There exist an element with order 2 since 2 is prime. Then $|[G,G]|=2$ $\endgroup$
    – S.O
    Apr 22 at 17:09
  • $\begingroup$ The subgroup generated by this element of order $2$ has no reason to be normal, so you can’t say that it must contain the derived subgroup. $\endgroup$
    – Mindlack
    Apr 22 at 17:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.