# A non abelian group of order 2p has 2 linear irreducible characters

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 *

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.

• Can't we do the same for 2. There exist an element with order 2 since 2 is prime. Then $|[G,G]|=2$
– S.O
Apr 22 at 17:09
• 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. Apr 22 at 17:35