7.30. Show that there are no simple groups of order 56.

Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter4: More On Groups
Section4.8: Some Results On Finite Abelian Groups (optional)
Problem 6E: 6. For each of the following values of , describe all the abelian groups of order , up to...
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Could you explain how to show 7.30 in detail? I also included lists of definitions and theorems in the book as a reference.

 
Theorem 7.11 (Cauchy's Theorem). Let G be a finite group, and suppose that a
prime p divides its order. Then G has an element of order p.
Corollary 7.3. Let p be a prime. Then a finite group G is a p-group if and only if
|G| = p", for some n > 0.
Corollary 7.4. Let p be a prime and G a finite group. Then G has just one Sylow
p-subgroup if and only if the Sylow p-subgroup is normal in G.
Theorem 7.12. Let G be a group of order pq, where p and q are primes with p < q.
Then the Sylow q-subgroup of G is normal. In particular, G is not simple.
Example 7.11. Let us show that there are no simple groups of order 351. As 351 =
33. 13, we note that the number of Sylow 3-subgroups is 1 + 3k and divides 13 and
the number of Sylow 13-subgroups is 1+ 131 and divides 27, with k, l e Z. The
only solutions for l are 0 and 2; that is, the number of Sylow 13-subgroups is either
1 or 27. If it is 1, then we know that the Sylow 13-subgroup is normal, and we are
done. So let us assume that it is 27. Now, each Sylow 13-subgroup is of order 13.
In a group of prime order, everything but the identity has order equal to that of the
group; thus, each Sylow 13-subgroup has 12 elements of order 13. Furthermore, if P
and Q are different Sylow 13-subgroups, then since |P N Q| must divide |P| = 13,
either P = Q (which is impossible) or P n Q = {e}. Thus, each of the 27 Sylow
13-subgroups contributes 12 elements of order 13, and there is no overlap. We have
now used up 12 · 27 = 324 elements of the group. This leaves only 351 – 324 = 27
elements. But that is the size of one Sylow 3-subgroup! Thus, there is only room
for one such subgroup. In order words, either the Sylow 13-subgroup or the Sylow
3-subgroup must be normal.
-
Theorem 7.13. Let G be a group of order pqr, where p, q and r are distinct primes.
Then G is not simple.
Theorem 7.14. Let G be a group of order p... p"*, where the p; are distinct
primes and the n¡ are positive integers. If, for each i, G has a unique Sylow
Pi-subgroup P;, then G = P¡ x ..× Pk.
Example 7.12. Suppose we wish to classify the groups of order 45. The number of
Sylow 3-subgroups is 1+ 3k and divides 5, for some k e Z. Thus, it can only be 1.
The number of Sylow 5-subgroups is 1+51 and divides 9, for some l e Z. Therefore,
the Sylow 5-subgroup is unique as well. According to the preceding theorem, a group
of order 45 must be the direct product of its Sylow subgroups. By Corollary 7.2, a
of order 9 is isomorphic to either Z3 x Z3 or Z,, and Corollary 4.2 tells us that
a group of order 5 is isomorphic to Z5. Hence, every group of order 45 is isomorphic
to either Z3 x Z3 × Z5 or Z9 × Z5 (and these are not isomorphic to each other, by
Theorem 5.6).
group
Transcribed Image Text:Theorem 7.11 (Cauchy's Theorem). Let G be a finite group, and suppose that a prime p divides its order. Then G has an element of order p. Corollary 7.3. Let p be a prime. Then a finite group G is a p-group if and only if |G| = p", for some n > 0. Corollary 7.4. Let p be a prime and G a finite group. Then G has just one Sylow p-subgroup if and only if the Sylow p-subgroup is normal in G. Theorem 7.12. Let G be a group of order pq, where p and q are primes with p < q. Then the Sylow q-subgroup of G is normal. In particular, G is not simple. Example 7.11. Let us show that there are no simple groups of order 351. As 351 = 33. 13, we note that the number of Sylow 3-subgroups is 1 + 3k and divides 13 and the number of Sylow 13-subgroups is 1+ 131 and divides 27, with k, l e Z. The only solutions for l are 0 and 2; that is, the number of Sylow 13-subgroups is either 1 or 27. If it is 1, then we know that the Sylow 13-subgroup is normal, and we are done. So let us assume that it is 27. Now, each Sylow 13-subgroup is of order 13. In a group of prime order, everything but the identity has order equal to that of the group; thus, each Sylow 13-subgroup has 12 elements of order 13. Furthermore, if P and Q are different Sylow 13-subgroups, then since |P N Q| must divide |P| = 13, either P = Q (which is impossible) or P n Q = {e}. Thus, each of the 27 Sylow 13-subgroups contributes 12 elements of order 13, and there is no overlap. We have now used up 12 · 27 = 324 elements of the group. This leaves only 351 – 324 = 27 elements. But that is the size of one Sylow 3-subgroup! Thus, there is only room for one such subgroup. In order words, either the Sylow 13-subgroup or the Sylow 3-subgroup must be normal. - Theorem 7.13. Let G be a group of order pqr, where p, q and r are distinct primes. Then G is not simple. Theorem 7.14. Let G be a group of order p... p"*, where the p; are distinct primes and the n¡ are positive integers. If, for each i, G has a unique Sylow Pi-subgroup P;, then G = P¡ x ..× Pk. Example 7.12. Suppose we wish to classify the groups of order 45. The number of Sylow 3-subgroups is 1+ 3k and divides 5, for some k e Z. Thus, it can only be 1. The number of Sylow 5-subgroups is 1+51 and divides 9, for some l e Z. Therefore, the Sylow 5-subgroup is unique as well. According to the preceding theorem, a group of order 45 must be the direct product of its Sylow subgroups. By Corollary 7.2, a of order 9 is isomorphic to either Z3 x Z3 or Z,, and Corollary 4.2 tells us that a group of order 5 is isomorphic to Z5. Hence, every group of order 45 is isomorphic to either Z3 x Z3 × Z5 or Z9 × Z5 (and these are not isomorphic to each other, by Theorem 5.6). group
7.30. Show that there are no simple groups of order 56.
2·3·5· 29. Show that at least one of the
7.32. Let G be a group of order 870
Sylow p-subgroups of G must be normal, for some prime p dividing |G|.
Transcribed Image Text:7.30. Show that there are no simple groups of order 56. 2·3·5· 29. Show that at least one of the 7.32. Let G be a group of order 870 Sylow p-subgroups of G must be normal, for some prime p dividing |G|.
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