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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!!

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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.

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  • $\begingroup$ 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

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