## Algebra Solutions: Hungerford II.5

1) If $N \triangleleft G$ and N, $G/N$ are both p-groups, then G is a p-group.
Proof-
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 $H\cap C(G) \neq \{e\}.$
Proof-
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 $p^k$.
Proof-
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 $\leq p^m$.
Proof-
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 $f(P) \leq P$.
Proof-
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.
Proof-
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 $S_3,S_4,S_5.$
Proof-
$|S_3| = 6 = 2 \cdot 3$. 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.
Proof-
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.
Proof-
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?
Proof-
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.
Proof-
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.