http://www.auburn.edu/~huanghu/math5310/alg-hw-ans-13 (I think).pdf

Apologies if I missed some backslashes which are induced by InftyReader version Does anyone know how to fix this?

14.31 Show: An intersection of normal subgroups of a group G is again a normal subgroup of G.

Solution: Let $H_{i}$ be normal subgroups of G (for $i \in I$ where $I$ is an\ index\ set). Then for\ $g\in G$ by left and right cancelation laws and the normality of $H_{i}$\ we\ have $g\bigcap_{i\in I}H_i = \bigcap_{\Large{i\in I}}(\color{red}gH_{i})=\bigcap_{i\in I}(H_{i}g)=(\bigcap_{i\in I}H_{i})\ \color{red}g\ . $
This shows $\bigcap_{i\in I}H_{i}$\ is a normal subgroup of $G$.

$\color{darkred}{ \text{ (1.) Why are you authorized to push $g$ into the intersection? Why are you authorized to take $g$ out? } }$

14.32. Given any subset S of a group G, show that it makes sense to speak of the smallest normal subgroup that contains S. [Hint: Use Exercise 31]

Solution: Let $\{H_{i}|i\in I\}$ be the collection of all normal\ subgroups of $G$ that contain $S$. Then by Exercise\ 31, the subset $ \bigcap_{i\in I}H_{i} \vartriangleleft G$. Obviously $ \bigcap_{i\in I}H_{i} $\ contains\ $S$.\ So $ \bigcap_{i\in I}H_{i} $\ is the smallest normal subgroup that contains $S$.

(2.) I think $\trianglelefteq$ means $\vartriangleleft$ or $=$. Indefectible? I understand $ \bigcap_{i\in I}H_{i} \trianglelefteq G$ and $\bigcap_{i\in I}H_{i} \subseteq H_i$.
But how does this proof flesh out or induce $\\bigcap_{i\in I}H_{i} \trianglelefteq G \;$ is the smallest $\;\trianglelefteq G$?

(3.) What's the intuition for (31.) and (32.)?

  • $\begingroup$ For (1), use the coset trick I mention in this answer to your other current post. For (2), yes your interpretation of the symbol is correct. Although don't use the word "perfect" here, as perfect subgroups are a thing... $\endgroup$ – user1729 Feb 11 '14 at 17:01

(1) We can show that $\bigcap_{i \in I} g H_i = g \bigcap_{i \in I} H_i$. Define $\bigcap_{i \in I} H_i = \{h \in G : h \in H_i \text{ for all } i \in I\}$. Now $$g\bigcap_{i \in I} H_i = \{gh \in G : h \in H_i \text{ for all } i \in I\}.$$ On the other hand, $$\bigcap_{i \in I} gH_i = \{h \in G : \color{brown}{h \in gH_i} \text{ for all } i \in I\}.$$

Proof for $\bigcap_{i \in I} g H_i \subset g \bigcap_{i \in I} H_i$:
If $\color{brown}{h \in gH_i} $ for all $i$ then $g^{-1}h \in H_i$ for all $i$. Therefore $g(g^{-1}h) = h$ is an element of $g \bigcap_{i \in I} H_i$.

If $h \in H_i$ for all $i$ then $gh \in gH_i$ for all $i$. This shows the other inclusion.

(2) You are right about that symbol. The proof does not explicitly argue that the intersection is the smallest such normal subgroup, but the argument is common in algebra and you should try to work it out. To show that $\bigcap_{i \in I} H_i$ is the smallest normal subgroup containing $S$, you need to show that any other normal subgroup containing $S$ is also contains that intersection. So suppose that $K$ is a normal subgroup which contains $S$. Then...

(3) Do you know about group actions? You can think of each element of $G$ as defining a homomorphism from $G$ to itself as follows: let $g \in G$. The function $\sigma_g: G \to G$ is defined by $\sigma_g(h) = g^{-1}hg$. So a subgroup $H$ is normal if $\sigma_g(H) = H$ for all $g \in G$. Such subgroups are "invariant under action by $G$" -- their elements might get moved, but the groups themselves stay the same.

So you can think of this dynamically: the elements of $G$ shuffle the subgroups around, but $H$ and $K$ are firmly rooted and will not be moved. So their intersection is also rooted.

  • $\begingroup$ Thanks a lot. Upvoted. Group actions are in 2 chapters. I'll get back here after. $\endgroup$ – Nai Feb 21 '14 at 9:33

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