0
$\begingroup$

I am trying to understand the definition of subdirect subgroup and full diagonal subgroup. But I am finding it hard to undertstand. Could anyone explain it with example? Particularly when we are to find the subdirect subgroup of a direct product of non abelian simple groups? For example $G=A_5 \times A_5 \times A_5$. What will be it's subdirect subgroup and diagonal subgroup? There is a lemma 1.4.1 in the thesis http://www.joannafawcett.com/masters.pdf on page 12.

I am particularly interested to understand the 2nd part of this lemma which states that Let $G = T_1 \times T_2 \times \cdots \times T_k$ be a direct product of nonabelian, simple groups $T_1,\ldots,T_k$ ($k ≥ 1$). Let $H$ be a subgroup of $G$ and $I := \{1,\ldots, k\}$. If $H$ is a subdirect subgroup of $G$, then $H$ is a direct product $\prod H_j$ , where each $H_j$ is a full diagonal subgroup of some subproduct $\prod_{i \in I_j} T_i$ and $I$ is partitioned by the $I_j$ .

I am looking for it's example. Because I couldn't make sense what does partition means here in the form of $T_i$. For example if we have $G=A_5 \times A_5 \ldots A_5$. Then according to this lemma how H, T_i and I_j look like? How do we construct it by using this lemma?

$\endgroup$
0
$\begingroup$

The diagonal subgroup of $A_5\times A_5\times A_5$ is given by $$ D=\{(g,g,g)\mid g\in A_5\}, $$ and for any epimorphism $\phi_i\colon A_5\rightarrow A_5$, the subgroup of $A_5\times A_5$ given by $\{(g,\phi(g))\mid g\in G\}$ is a subdirect product of $A_5$ and $A_5$. Similarly for $A_5\times A_5\times A_5$. The definition here is good to understand, I think.

$\endgroup$
  • $\begingroup$ Thanks Dietrich, but could you tell me please that Subdirect subgroup and subdirect product are same thing? if not what's the difference? $\endgroup$ – S786 Mar 12 '17 at 20:17
  • $\begingroup$ joannafawcett.com/masters.pdf Trying to understand part 2 of lemma 1.4.1 on page 12. $\endgroup$ – S786 Mar 12 '17 at 20:18
  • $\begingroup$ Well, the section before Lemma $1.4.1$ answers your question on subdirect products and subdirect subgroups - see line $5$. $\endgroup$ – Dietrich Burde Mar 12 '17 at 21:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.