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Let and in an element of order for some.
If p = 2, 3, then G is a group of order six or twelve and not an element of order six.
In this case, we have 3 ∈ π ( H ), so an element of order 5 must act fixed-point-freely on a subgroup of order 3 in H, which is clearly a contradiction by Table 2.
Regarding other cases of | G / K |, we always can find an odd prime r ∈ π ( H ) and r | p − 1 2 such that ( p, | Aut ( H r ) | ) = 1, which implies that G has an element of order pr, a contradiction.
Then there exists an odd prime r ∈ π ( ( p + 1 ) / 2 ) such that | H r | < p. By Lemma 2.5, ( p, | Aut ( H r ) | ) = 1, which implies that an element of order p of G can trivially act on H r. In other words, r can be connected to p in the prime graph of G, a contradiction.
In 1970s, a simple graph called prime graph of the group G was introduced: the vertex set of this graph is π ( G ), two vertices p and q are joined by an edge if and only if G contains an element of order pq (see [1]).
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That is, before there was a historical Jesus born of Mary and accessible to the sight and touch of Jews and others in his own day, there was a Logos a principle of reason, an element of ordering, a "word"—that participated in the Godhead and thus existed, but which only preexisted as far as the "incarnate" Logos, the word that took on flesh and humanity (John 1 1 14), was concerned.
Prior to use of the protocol, the Identity Provider must select three domain parameters,,, and, where and are large primes satisfying, and is an element of multiplicative order in.
Theorem 12 If H u, n contains a triangle, then Γ 1 ( n ) contains an elliptic element of order 3. Proof If H u, n contains a triangle, then by Theorem 9, n = 1.
Conversely, assume that Γ 1 ( n ) contains an elliptic element of order 2. Then there is an element of Γ 1 ( n ) of the form ( 1 + a n b c n 1 + d n ) such that 2 + ( a + d ) n = 0. From this we get n = 1 or 2. Hence, the proof now follows from Theorem 10.
For example, the element ( 1 − 1 3 − 2 ) ∈ Γ 1 ( 3 ) is an elliptic element of order 3, but H 1, 3 does not contain a triangle.
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Since I tried Ludwig back in 2017, I have been constantly using it in both editing and translation. Ever since, I suggest it to my translators at ProSciEditing.

Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com