Algebra Solutions: Hungerford II.5

1) If N \triangleleft G and N, G/N are both p-groups, then G is a p-group.
Since N and G/N are p-groups let |N| = p^i, |G/N| = p^j for some i and j. Then by Lagrange we have |G| = [G:N]|N| = |G/N| |N| = p^ip^j = p^{i+j}. Thus G is a p-group.

2) If G is a finite p-group, H\triangleleft G and H is nontrivial, then H\cap C(G) \neq \{e\}.
Note that the center of G is C(G) = \{ g\in G| gx = xg \forall x \in G\}. Let G be a finite p-group so that |G| = p^n. Then since H is a subgroup we have |H| = p^m where m \leq n. We can let G act on H by conjugation and let H_0 = \{h\in H | ghg^{-1} = h \forall g \in G\}. Then we have by theorem that |H|\equiv |H_0| \pmod{p}. We can also write H_0 = \{ h\in H | gh = hg \forall g\in G\}. Now if a\in H_0 \Rightarrow a\in H, and ga = ag \forall g\in G \Rightarrow a\in H \cap C(G). Therefore H_0 \subset H \cap C(G). We also have that p divides the order of H_0 so that 1 < |H_0| \leq |H \cap C(G)|. Thus H\cap C(G) must have more than one element and is therefore nontrivial.

3) Let |G| = p^n. For each k, 0 \leq k\leq n, G has a normal subgroup of order p^k.
First note that any subgroup of the center of a group is automatically normal in the group. Now we prove that G has a normal subgroup of order p. Since G is a p-group it has a nontrivial center so that the center is in fact a p-group as well. Then being of order a prime power, we can say by Cauchy’s Theorem that there exists g \in C(G) such that |g| = p. Then \langle g \rangle is a subgroup of the center of G, so is automatically normal in G, and has order p.
Now we show for each k, 0 \leq k\leq n, G has a normal subgroup of order p^k by induction on k. If k = 0 we have the trivial subgroup which is automatically normal in G. Now assume that G has normal subgroups for each p^i where 0 \leq i \leq k and call them N_i. Consider the quotient group G/N_k. We have from Lagrange that |G/N_k| = p^{n-k} so that G/N_k is a p-group. In the previous step we’ve shown that a p-group must contain a normal subgroup of order p, call it \langle aN_k \rangle . Let \pi : G \rightarrow G/N_k be the canonical projection map. Then let H = \pi^{-1} (\langle aN_k \rangle ) = \{ g\in G| gN_k \in \langle aN_k \rangle \}. The preimage of a normal subgroup under a homomorphism is also normal, so that H is normal in G. Note that N_k \leq H because N_k is the identity in G/ N_k so that N_k \in \langle aN_k \rangle so that \pi^{-1} (\{N_k \}) = N_k \leq \pi^{-1}(\langle aN_k \rangle )= H . Then using Lagrange we have |H| = [H:N_k] |N_k|, but H/N_k = \langle aN_k \rangle \Rightarrow  |H/N_k| = |\langle aN_k \rangle | = p . Therefore |H| = p \cdot p^{k} = p^{k+1}

4) If G is an infinite p-group, then either G has a subgroup of order p^n for each n\geq 1 or there exists m\in \mathbb{N} such that every finite subgroup of G has order \leq p^m.
Suppose that no such m exists. Then for each positive integer n, there exists another positive integer N>n such that G has a subgroup H of order p^N. But then by Sylow we know that H has a subgroup of order p^n.

5) If P is a normal Sylow p-subgroup of a finite group G and f: P \rightarrow P is an endomorphism, then f(P) \leq P.
Note that f(P) \leq G. Let y\in f(P). Then there exists a x\in P such that f(x)= y. Since P is a p-group, |x| = p^i for some i.Then we have y^{p^i} = (f(x))^{p^i}) = f(x^{p^i}) = f(e) = e. This tells us that |y| divides p^i so that |y| = p^j for some j\leq i. Since y was arbitrary, f(P) is a p-subgroup of G. Therefore by Sylow’s Second Theorem, there exists x\in G such that f(P) \leq xPx^{-1} = P.

6) If H is a normal subgroup of order p^k of a finite group G, then H is contained in every Sylow p-subgroup of G.
H is a p-subgroup of a finite group G. Let P be any Sylow p-subgroup of G. From Sylow’s Second Theorem we know there exists x\in G such that H \leq xPx^{-1}. We also have that x^{-1}Hx is nonempty. Let y\in x^{-1}Hx. Then we can write y = x^{-1}hx for some h\in H. Since H\leq xPx^{-1} for all h\in H, we can write h = xpx^{-1} for some p\in P. Then rewrite y = x^{-1}xpx^{-1}x = p\in P. Therefore x^{-1}Hx \subset P. Lastly, let a,b \in x^{-1}Hx. Then we can write a=x^{-1}h_1x, b = x^{-1}h_2x for some h_1,h_2 \in H. Then we have ab^{-1} = x^{-1}h_1xx^{-1}h_2^{-1}x = x^{-1}h_1h_2^{-1}x \in x^{-1}Hx so that in fact x^{-1}Hx is a subgroup of P. Now, since H is normal in G we can say that H = xHx^{-1} = x^{-1}Hx \leq P, so that H is contained in any arbitrary Sylow p-subgroup P.

7) Find the Sylow 2-subgroups and Sylow 3-subgroups of S_3,S_4,S_5.
|S_3| = 6 = 2 \cdot 3. Thus S_3 contains at least one Sylow 2 subgroup of order 2 and one Sylow 3 subgroup of order 3 by the first Sylow Theorem. In fact, from Sylow’s Third Theorem we know that there is either 1 Sylow 2 subgroup or 3 Sylow 2 subgroups. We also know that there can only be one Sylow 3 subgroup.
In the same manner we can say that S_4 has either 1 or 3 Sylow 2 subgroups of order 8 and either 1 or 4 Sylow 3-subgroups of order 3. This is because |S_4| = 12 = 2^3 \cdot 3.
Lastly, S_5 has either 1,3,4, or 15 Sylow 2 subgroups of order 8 and either 1,4, or 10 Sylow 3-subgroups of order 3. This is because |S_5| = 120 = 2^3 \cdot 3 \cdot 5.

8) If every Sylow p-subgroup of a finite group G is normal for every prime p, then G is the direct product of its Sylow subgroups.
First we write |G| = p_1^{r_1}p_2^{r_2}\cdots p_1^{r_n}. where we have distinct primes. Then by Sylows first theorem, for each prime p_i, G contains a Sylow p_i subgroup of order p_i^{r_i}. From the second Sylow theorem we have that any two Sylow p_i subgroups are conjugate, but we are given that they are all normal so in fact there is a unique Sylow p_i subgroup for each p_i. Denote each Sylow p_i subgroup by H_i. Then we have that \{H_i\}_{i=1}^n is a family of normal subgroups of G so that \langle \cup_{i=1}^m H_i \rangle = H_1 \cdots H_m for any 1\leq m\leq n.
Next we need to show that for each j\in \{1,\ldots ,n\} that H_j \cap \langle \cup_{i\neq j} \rangle = \{e\}, for all n\in \mathbb{N}. This is done by induction on n. Lastly we have to show that G = \langle \cup_{i=1}^n H_i \rangle , but we already have that H_1 \cdots H_n = \langle \cup_{i=1}^n H_i \rangle \leq G so that we need only show that | H_1 \cdots H_n| = |G|. Then we will have that G \cong H_1 \times \cdots \times H_n.

9) If |G| = p^nq with p>q primes, then G contains a unique normal subgroup of index q.
From Sylow’s First Theorem, G has a Sylow p-subgroup of order p^n. Let m be the number of Sylow p subgroups of G. By Sylows Third Theorem we have that m | |G| and m\equiv 1 \pmod{p}. Then since |G| = p^nq we have to have that either m=1,q or p^i for some 1\leq i \leq n. If m = q then p|(q-1) \Rightarrow q-1 = pk for some k\in \mathbb{Z}. Then pk+1 = q < p \Rightarrow k = 0 \Rightarrow q =1 which is a contradiction. If instead we have that m = p^i then p | (p^i-1) \Rightarrow p = 1, which is another contradiction. Therefore m =1 so that there is the only 1 Sylow p subgroup of G, call it H, which must be normal in G. Then by Lagrange, |G| =  [G:H]|H| \Rightarrow p^nq = [G:H] p^k \Rightarrow q = [G:H]. Thus H has index q.

10) Every group of order 12, 28, 56, and 200 must contain a normal Sylow subgroup and hence is not simple.
Proof- i)
|G| = 12 = 2^2\cdot 3. By 3rd Sylow Theorem we know that there is either 1 Sylow 3 subgroup, or 4 Sylow 3 subgroups. If there is only one, then know it must be normal. Thus, suppose there are 4, H_1,H_2,H_3,H_4 each with order 3. Each of these subgroups has 3 elements, 1 being the identity. Thus the other 2 elements are of order 3. Together then we have 4(3-1) = 8 elements of order 3 in G. Then G can only have one Sylow 2-subgroup which must be of order 4 because this gives us 12 total elements. Again, being the unique Sylow 2-subgroup gives that it is also normal.

ii) |G| = 28 = 2^2 \cdot 7. The Third Sylow Theorem tells us that the number of Sylow 7 subgroups is either 1 or 4, and that it must also be 1 mod 7, therefore it cannot be 4. So there is only one Sylow 7-subgroup and thus it is normal.

iii) |G| = 56 = 2^3 \cdot 7. By the Third Sylow Theorem we know that the number of Sylow 7-subgroups is either 1 or 8. If it is 1 then we are done, so suppose it is 8. Call them H_1, H_2,\ldots , H_8. Then similar to part (i) we have that there are 8(7-1) = 48 elements of order 7 in G. Then G can only have one Sylow 2-subgroup as it must be of order 2^3 = 8. Therefore it is normal, being unique.

iv) |G| = 200 = 2^35^2. By the Third Sylow Theorem we have that the number of Sylow 5-subgroups is a factor of 2^3 or equal to 1, and must also be congruent to 1 modulo 5. This can only happen if there is exactly one Sylow 5-subgroup. Thus it is normal.

11) How many elements of order 7 are there in a simple group of order 168?
Let |G| = 168 = 2^3\cdot 3 \cdot 7. By the Third Sylow Theorem, the number of Sylow 7-subgroups must divide 2^3\cdot 3 and also must be congruent to 1 modulo 7. Thus the only possibilities are that the number of Sylow 7-subgroups is 1 or 8, but since G is simple we cannot have only one Sylow 7-subgroup otherwise it would be normal. Therefore there are 8 of them, each with 7 elements, with each non identity element having order 7 since a group with prime order must be cyclic. Thus there are 8(7-1)=48 elements of order 7.

13) Every group G of order p^2 is abelian.
Since G is a p-group, the center C(G) \neq \{e\}. By Lagrange we then have either |C(G)| = p^2 which gives the center as the entire group, thus making G abelian, or we have |C(G)| = p. If |C(G)| = p then we have that G/C(G) is cyclic being of prime order, which tells us that G is abelian by 2.4.9.

This entry was posted in Algebra Solutions. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s