Everything posted by Genady
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Is H a normal subgroup...
I think so.
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Is H a normal subgroup...
Nice. So, you've proved that a simple group cannot have subgroups of order half that of the group. 😉 Because I know the answer.
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Is H a normal subgroup...
... if it's a subgroup of a finite group G and contains exactly half the elements of G?
- Locked Books
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Locked Books
Perhaps. But I don't see this number in the text.
- Locked Books
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How are the orders in a cyclic group related?
It is also sufficient, right?
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How are the orders in a cyclic group related?
"It can be also generated by its other element" is a given.
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How are the orders in a cyclic group related?
G is a cyclic group of order n generated by an element g, G = {e, g, g2, ..., gn-1). It can be also generated by its other element h = gk, G = {e, h, h2, ..., hn-1}. How are the k and n related?
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Do equivalence classes form a partition?
I think I'm missing something. If g1 = h1 g h1-1 and g2 = h2 g h2-1 then g1 ~ g and g2 ~ g, but it is not necessary that g1 ~ g2, which they should if this is an equivalence class. ?
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Azerbaijan Airlines plane shot down by Russian missile on Xmas Day
As Baku native, I could be offended (I am not.)
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Do equivalence classes form a partition?
Does it mean, g1 ~ g when for some h1, g1 = h1 g h1-1 ?
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Is being closed a sufficient condition for a finite subset to be a subgroup?
IOW, Let's take any h in H. Because of H being closed, all combinations h, hh, hhh, ... are in H. But because of H being finite, some combinations should repeat, say hm = hn for some m<n. Then hn = hmhn-m = hm and thus hn-m is identity, e (from G). e = hn-m is thus contained in H. Now, if n-m = 1, then e = h is inverse of itself. Otherwise, e = hhn-m-1, which makes hn-m-1 an inverse of h.
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Is being closed a sufficient condition for a finite subset to be a subgroup?
I think, that being finite and closed guarantees that it contains the identity and all the inverses.
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Is being closed a sufficient condition for a finite subset to be a subgroup?
H is a finite subset of a group G, and for each h1, h2 in H their combination h1h2 is also in H. Is H a group?
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Do equivalence classes form a partition?
Plus, even without other equivalent elements, each element forms a class of itself.
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Plant biology are underestimate (podcast idea)
OK then.
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Plant biology are underestimate (podcast idea)
Have you noticed that you reply to a four years old post to a member who did not visit since September 2021?
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Do equivalence classes form a partition?
Equivalence relation R on set X defines equivalence classes in X. Do they form a partition of X?
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Is it in the interest of SSA to delay processing of applications for benefits?
I have applied for my retirement benefits in November. My application is still in process. I understand that eventually they will pay retroactively whatever accumulates since November. Without any interest. Thus, the OP question.
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The myth of invasive lionfish
^^^^ fact, but ^^^^ speculation.
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The myth of invasive lionfish
They have, e.g., Invasive Cup Coral | Flower Garden Banks National Marine Sanctuary
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What are you listening to right now?
When the first plane crashed, I was straight under the WTC. 500m was how far I walked away before the structures collapsed.
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What are you listening to right now?
It sure did. I watched it from about 500 m.
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The myth of invasive lionfish
Almost three years after the OP, here is the status. One of the main lionfish hunters got bends while hunting lionfish and left the island long ago together with her partner, the other lionfish hunting champion. Since COVID, lionfish hunting practically has stopped. It is not promoted anymore. The result: no effect whatsoever. The lionfish is there, there is no more and no less of it, as well as the rest of the reef. Evidently, the devastating effect that has been predicted, was in fact a myth.