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In the following image (from "Field Arithmetic by Fried & Jarden" Page 6, Lemma 1.2.2(b)), red rectangle,

enter image description here

I'm trying to figure out why it's right to claim $h^{-1} \in H$.
I thought the following solved it:

$g=k_ih_i$ and $g=kh^{-1}$ $\Rightarrow k_ih_i=kh^{-1} \Rightarrow h_ih=k_i^{-1}k \in H$ because right hand side is in H.
So $h^{-1}=k^{-1}g=k^{-1}k_ih_i$, hence, because $k_i^{-1}k \in H$ we get $h^{-1} \in H \Rightarrow g=kh^{-1} \in KH$.

But then I realized there's no logic in assuming $h_ih \in H$ because we're talking subsets here, not subgroups. So is it ok to assume $h_ih \in H$? if so, why?
If not, I'd appreciate an explanation for the red part..

Khal
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  • The symbol $\leq$ is usually reserved for subgroups. It looks to me that they are asserting that $H_i$ is a subgroup of the finite intersection. – Arturo Magidin May 09 '19 at 17:14
  • P.S. It’s generally preferable not to post images, but rather to copy the text; images cannot be searched, and there are other interface issues with them. – Arturo Magidin May 09 '19 at 17:15
  • Appologies, I didn't know that.. – Khal May 09 '19 at 17:20
  • As I commented to user672573 below, $H_is$ are defined as subsets, not subgroups – Khal May 09 '19 at 17:25
  • I know, I can read. But later on, they use the symbol $\leq$, which, as I indicated, is normally reserved for subgroups; the property seems to suggest that your assumptions include that the $H_i$ are also subgroups. Otherwise, propery (1) would read “$H_i\subseteq \cap_{j\in J} H_j$” rather than “$H_i\leq \cap_{j\in J} H_j$”. Which I thought was clear from the syntax of my sentence. – Arturo Magidin May 09 '19 at 17:37
  • I get you, however, even if we assume that's correct, it relates only to a finite subset J of I, while the proof refers to any i in I. – Khal May 09 '19 at 17:43
  • And if $J={i}$, it follows that you either get $H_i$ is a subgroup, or else that there is an $H_{i’}$ contained in $H_i$ which is a subgroup, so if you replace $H$ with the intersection of all $H_i$ which are subgroups, you get the exact same intersection. – Arturo Magidin May 09 '19 at 17:55
  • You're assuming that this method will cover all $H_is$ but in fact, it can be that for each $J$ as you mentioned, the subgroup will be only H_1 and so the intersection will contain only H_1 as oppose to all $H_is$ as required. – Khal May 09 '19 at 18:42
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    If you always get $H_1$, then you get that $H_1$ is contained in $H_j$ for all $j$, so that the intersection would actually be contained in all $H_j$. Look: define $H$ to be the intersection over all $i$ for which $H_i$ is actually a subgroup. I claim that $H$ is contained in the intersection of all $H_j$. To prove that, let $j\in I$ be fixed. Then by the condition given, there exists $i\in I$ such that $H_i$ is a subgroup of $\cap_{j\in{j}}H_j = H_j$. Thus, $H\subseteq H_i\leq H_j$. Hence, $H\subseteq H_j$. This proves the claim. – Arturo Magidin May 09 '19 at 18:51
  • In other words, you conditions guarantees that you have a cofinal subset of $I$ that indexes subgroups. Since the limit over a cofinal subset is equal to the whole limit, you can replace your original set with the set of all $H_i$ that are subgroups and get the exact same result. – Arturo Magidin May 09 '19 at 18:53
  • @ArturoMagidin I agree to you concerning the use of $\le$. However, the result is true for arbitrary closed $H_i$ and $\subseteq$. The proof contains a silly mistake. See my answer. – Paul Frost May 10 '19 at 12:53
  • @ArturoMagidin I agree (using your notations) that $H \subseteq H_i \le H_j$ as you wrote. what makes you say that it's actually $H=\bigcap_{j \in I}H_j$? There can still be $h \in \bigcap_{j \in I}H_j$ s.t. $h /\in H$.. what am I missing? – Khal May 10 '19 at 13:40
  • I don't understand what your issue is, and the comment thread is way too long. If you replace your family with the family of $H_i$ that are actually subgroups, you get the exact same intersection and the argument, if correct, would go through. Your objections continually miss the fact that, under the assumption I understood to be happening, you can replace any $H_j$ with a smaller $H_{j'}$ which is a subgroup, so that you can argue with subgroups throughout. Though as Paul Frost noted, that is immaterial. – Arturo Magidin May 10 '19 at 15:48

2 Answers2

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The proof is not correct.

You have $g = k_ih_i$ with $k_i \in K, h_i \in H_i$. But then $h_i = k_i^{-1}g \in K^{-1}g$ and we conclude $H_i \cap K^{-1}g \ne \emptyset$ and not $H_i \cap g^{-1}K \ne \emptyset$. But then we can correctly show that there exists $h \in \bigcap_{i \in I} (H_i \cap K^{-1}g) = (\bigcap_{i \in I} H_i) \cap K^{-1}g = H \cap K^{-1}g$. Hence $h \in H$ and $h = k^{-1}g$ for some $k \in K$. This means $g = kh \in KH$.

Paul Frost
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  • Seems legit. Thanks!
    What can you say about the $\le$? I'm still not sure if it's meant to show $H_i$ is a subgroup of the intersection or if it's just the binary relation $\subseteq$ that comes from the "directedness" of the family of subsets.     – Khal May 10 '19 at 13:12
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    I believe Arturo Magidin is right: $\le$ is a relation for subgroups. But I do not know your book, perhaps the author uses $\le$ also for subsets or perhaps it is a typo. Anway, the result is correct for $\subseteq$. – Paul Frost May 10 '19 at 13:16
  • But if $A \le B$ is really intended to denote the subgroup relation, then it only makes sense to use it if $B$ is a group and $A$ is a subgroup. This would implicitly say that all $H_i$ are subgroups (take $J = { i }$). This would not fit to the assumption the $H_i$ are subsets. – Paul Frost May 10 '19 at 13:28
  • That I can't understand. Why would that imply $H_i$ are all subgroups? – Khal May 10 '19 at 13:42
  • Take $J ={ i' }$. The assumption implies that there is $i$ such that $H_i \le \bigcap_{j \in J} H_j = H_{i'}$. Hence $H_{i'}$ must be a subgroup if you use $\le$ "correctly". – Paul Frost May 10 '19 at 13:47
  • Oh and by that $H_{i'}$ would have to be a group.. got it :) – Khal May 10 '19 at 13:51
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    I had a look into your book. On the preceeding pages the authors use $\subseteq$ for set inclusion. On p..5 $\le$ occurs for (closed normal) subgroups, this indicates that it is in fact used as the subgroup relation. Also part (a) of the Lemma gives a hint that the $H_i$ are intended to be subgroups. – Paul Frost May 10 '19 at 13:56
  • So by our recently acquired logic, each $H_i$ (element of the intersection) is in fact a subgroup itself and then it's ok to assume they're closed under inversion. Thanks for the help and patience @Paul Frost! – Khal May 10 '19 at 14:02
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    But note that the correct proof of (b) does not require that $H$ is closed under inversion ;-) We only need to know that $K^{-1}$ is closed which is due to the continuity of the inversion. – Paul Frost May 10 '19 at 14:05
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H is as an intersection of subgroups a subgroup by itself. As you can see almost immediatly there has to exist an invert in H for every element of H. It follows the existence of $h^{-1}$ in H.

  • But $H_is$ are defined as subsets, not subgroups.. – Khal May 09 '19 at 17:24
  • Look here. link @Shmooze –  May 09 '19 at 17:29
  • Thanks, but the question in the link assumes H is closed under the product of G, while in my case, they're just closed.. – Khal May 09 '19 at 17:38
  • H is closed under group action in your case. It means the same as closed under the product of G. –  May 09 '19 at 17:46
  • What makes you say H is closed under group action here? – Khal May 09 '19 at 17:50
  • Because of the title you choosed I assume closed means closed in terms of group actions here. –  May 09 '19 at 17:56
  • The title states that the subset is not closed under group action – Khal May 09 '19 at 18:05
  • Then you've to say us what closed means in this context otherwise we won't be able to help. It's a word used many times in the mathematics with many different meanings, but closed in terms of group action would make the most sense in my opinion. –  May 09 '19 at 18:25
  • I'm sorry but the question states specifically that the word closed in this context refers to the subset and not the group action. The word "under" didn't appear anywhere but the title and there it specifically said "Subset of a subgroup is not closed under group actions". Sorry if you were mislead, that wasn't my intention. – Khal May 09 '19 at 18:32