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Let $G$ be a finite group. It is stated that

group $G$ is sovable $\iff$ there exists a normal series $$\{e\}=H_s \lhd H_{s-1} \lhd \cdots \lhd H_{1} \lhd H_0 = G$$ such that each quotient group $H_i/H_{i+1}$ is abelian $p_i$-group and each $H_i$ is normal in $G$.

Attempted solution:

"$\Longleftarrow$". Suppose that the normal series with above mentioned property exists. Then, using the fact that $G^{(1)}=[G,G]$ is the smallest normal subgroup with abelian quotient, we have $G^{(1)}\leq H_{1}$. This implies that $G^{(2)}=[G^{(1)},G^{(1)}] \leq [H_{1},H_{1}]\leq H_{2}$ and so on, so we have $G^{(i)} \leq H_{i}$ and hence $G^{(s)} = \{e\}$ which means that $G$ is solvable.

"$\Longrightarrow$". Now let $G$ be a solvable group. Then we have a series $$ \{e\}=G^{(k)} \lhd G^{(k-1)} \lhd \cdots \lhd G^{(1)} \lhd G^{(0)} = G $$ and $G^{(i)}/G^{(i+1)}$ is abelian and hence $$ G^{(i)}/G^{(i+1)} = \mathbb{Z}_{p_1^{m_1}} \oplus \ldots \oplus \mathbb{Z}_{p_s^{m_s}} $$ due to the fundamental theorem of finitely generated abelian groups (finite, in this case). So my guess is that we have to insert some subgroups $Y_1, \ldots, Y_{s-1}$ between $G^{(i)}=Y_0$ and $G^{(i+1)}=Y_s$ in such manner that $Y_j/Y_{j+1} = \mathbb{Z}_{p_j^{m_j}}$, but I'm not sure how to do it properly, so that's my question. Thank you in advance!

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A solvable group is one in which there is a sub-normal series $$G=N_0\geq N_1\geq N_2\geq\cdots\geq N_k=\{1\}$$ such that $N_i/N_{i+1}$ is abelian. (Sub-normal means $N_{i+1}$ is normal in $N_i$, but not necessarily in the whole group. Normal series usually refer to series in which each $N_i\lhd G$)

Now, given this definition, $\Leftarrow$ is clear.

Your guess in the other direction is correct. For each $i$, you have $N_i/N_{i+1}\cong\bigoplus_{j=1}^{M_i}\mathbb{Z}_{p_j}^{n_j}$. We therefore have a series $$N_i/N_{i+1}=L_{i,0}\geq L_{i,1}\geq\cdots\geq L_{i,M_i}=\{N_{i+1}\}$$ Where $L_k=\bigoplus_{j=1}^{M_i-k}\mathbb{Z}_{p_j}^{n_j}$. Using the correspondence theorem, you get a series of subgroups $$N_i= \tilde{L}_{i,0}\geq\tilde{L}_{i,1}\geq\cdots\geq\tilde{L}_{i,M_i}=N_{i+1}$$ such that $\tilde{L}_{i,k}/N_{i+1}=L_{i,k}$.

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Let $G$ be a group, $N$ a normal subgroup of $G$, and $q: G\to G/N$ the canonical quotient. If $K$ is a normal subgroup of $G/N$, then $q^{-1}(K)$ is a normal subgroup of $G$ such that $N \leq q^{-1}(K)$ and $q^{-1}(K)/N = K$.

Now apply this when $G=G^{(i)}$ and $N=G^{(i+1)}$ and $K= \mathbb{Z}_{p_1^{m_1}}$ to obtain $Y_1$ etc...

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