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Let $G$ be a $\{2,3\}$-group and $\lvert G\rvert=2^\alpha\cdot3^\beta$. For $p\in\{2,3\}$, define $$ \nu_p(G)\mathrel{:=}\min\left\{\log_p\left(\frac{\lvert G\rvert}{\chi(1)}\right)_p \mathrel{\bigg\vert} \chi\in\operatorname{Irr}(G)\right\}, $$ where $\operatorname{Irr}(G)$ is the set of all irreducible $\mathbb{C}$-characters of $G$.

Suppose that $\nu_2(G)=1$, $\nu_3(G)=0$. We want to study this group $G$. Especially, we want to know the answer to:

QUESTION: Are there two numbers $M$ and $N$ such that $\alpha<M$ and $\beta<N$ for all such groups $G$?

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  • $\begingroup$ For your question literally as stated one can take $M = \alpha + 1$ and $N = \beta + 1$, so I assume you meant "for all such groups $G$". I edited accordingly. $\endgroup$
    – LSpice
    May 14 at 20:19
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If I understand your conditions correctly, you are assuming that there is a $3$-block of $G$ of defect zero (this corresponds to an irreducible character $\chi$ with $\chi(1)_{3} = |G|_{3}$) and a $2$-block of $G$ of defect $1$ (for in general if a finite group $G$ has order divisibly by $p^{a}$ (but by no higher power of the prime $p$) and $G$ has an irreducible character $\chi$ of degree exactly divisible by $p^{a-1}$ (but by no higher power of $p$), then it is a result of R. Brauer that $\chi$ lies in a $p$-block of defect $1$.

Now since $|G| = 2^{\alpha}3^{\beta}$, we know that $G$ is solvable by Burnside's $p^{a}q^{b}$-theorem. Hence $C_{G}(F(G)) \leq F(G)$, where $F(G)$ is the unique largest nilpotent normal subgroup of $G$.

But for any prime divisor $p$ of $|G|$, $O_{p}(G)$ is contained in each defect group of each $p$-block of $G$, where $O_{p}(G)$ is the largest normal $p$-subgroup of $G$.

Hence we have $O_{3}(G) = 1$ and $|O_{2}(G)| \leq 2$. Since $G$ is a $\{2,3\}$-group, this implies that $F(G)$ has order dividing $2$, so that $F(G) \leq Z(G).$ Since $C_{G}(F(G)) \leq F(G)$, this forces $G = F(G)$ and $|G| \in \{1,2\}$. So the answer to your question is a resounding "yes".

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