Aratati ca pentru orice grup abelian finit \( G \) avem ca \( |End(G)|/|G| \) este patrat perfect.
(M. Andronache, I. Savu, GMA 1991)
Grupuri abeliene finite si patrate perfecte
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Dragos Fratila
- Newton
- Posts: 313
- Joined: Thu Oct 04, 2007 10:04 pm
Din teorema de structura a grupurilor abeliene finit generate rezulta ca
\( G\approx \mathbb{Z}_{d_1}\times\cdots\times\mathbb{Z}_{d_r} \) cu \( d_1|d_2|\ldots d_r \) si \( d_1d_2\cdots d_r=n=|G| \).
\( End(G) \) se identifica cu \( Hom_{\mathbb Z}(G,G) \), pt. ca \( G \) fiind abelian este un \( \mathbb{Z} \)-modul.
Din proprietatea de universalitate a produsului si sumei directe rezulta
\( Hom_{\mathbb{Z}}(\prod_{i=1}^r\mathbb{Z}_{d_i},\prod_{i=1}^r\mathbb{Z}_{d_i}) \approx \prod_{i,j}Hom_{\mathbb{Z}}(\mathbb{Z}_{d_i},\mathbb{Z}_{d_j}) \).
Se observa usor ca \( |Hom_{\mathbb{Z}}(\mathbb{Z}_{d_i},\mathbb{Z}_{d_j})|=\min(d_i,d_j) \)
Asadar
\( |End(G)|=\prod_{i,j}\min(d_i,d_j)=\prod_{1\le i\le j\le r}\min(d_i,d_j)\prod_{1\le j<i\le r}min(d_i,d_j)= \) \( \prod_{i=1}^rd_i^{r-i+1}\prod_{i=1}^{r-1}d_i^{r-i}
=d_r\prod_{i=1}^{r-1}d_i^{2(r-i)+1}=\left(\prod_{i=1}^{r-1}d_i^{r-i}\right)^2d_1d_2\cdots d_r \).
Cum \( |G|=d_1\cdots d_r \) rezulta concluzia.
\( G\approx \mathbb{Z}_{d_1}\times\cdots\times\mathbb{Z}_{d_r} \) cu \( d_1|d_2|\ldots d_r \) si \( d_1d_2\cdots d_r=n=|G| \).
\( End(G) \) se identifica cu \( Hom_{\mathbb Z}(G,G) \), pt. ca \( G \) fiind abelian este un \( \mathbb{Z} \)-modul.
Din proprietatea de universalitate a produsului si sumei directe rezulta
\( Hom_{\mathbb{Z}}(\prod_{i=1}^r\mathbb{Z}_{d_i},\prod_{i=1}^r\mathbb{Z}_{d_i}) \approx \prod_{i,j}Hom_{\mathbb{Z}}(\mathbb{Z}_{d_i},\mathbb{Z}_{d_j}) \).
Se observa usor ca \( |Hom_{\mathbb{Z}}(\mathbb{Z}_{d_i},\mathbb{Z}_{d_j})|=\min(d_i,d_j) \)
Asadar
\( |End(G)|=\prod_{i,j}\min(d_i,d_j)=\prod_{1\le i\le j\le r}\min(d_i,d_j)\prod_{1\le j<i\le r}min(d_i,d_j)= \) \( \prod_{i=1}^rd_i^{r-i+1}\prod_{i=1}^{r-1}d_i^{r-i}
=d_r\prod_{i=1}^{r-1}d_i^{2(r-i)+1}=\left(\prod_{i=1}^{r-1}d_i^{r-i}\right)^2d_1d_2\cdots d_r \).
Cum \( |G|=d_1\cdots d_r \) rezulta concluzia.
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