Fall 1994 Algebra Prelim Solutions

1.

(i)

By definition $\mathfrak {C}$(x) = {gxg^-1 | g $\in$ G} . Since x $\in$ H $\vartriangleleft$ G, we see that gxg^-1 $\in$ H for all g $\in$ G and we deduce that $\mathfrak {C}$(x) $\subseteq$ H. Let C_G(x) denote the centralizer of x in G. Then |$\mathfrak {C}$(x)| = [G : C_G(x)]. This last quantity is a power of p and it cannot be 1 because x is not in the center of G. We deduce that |$\mathfrak {C}$(x)| is a nontrivial power of p and (i) follows.

(ii)

It follows from (i) that H $\setminus$ Z is a union of conjugacy classes of the form $\mathfrak {C}$(x) where x $\in$ H $\setminus$ Z. Since different conjugacy classes intersect trivially, we see from (i) that p divides | H $\setminus$ Z|. Also | H| divides | G| and so | H| is a nontrivial power of p. We deduce that p divides | Z $\cap$ H| and (ii) is proven.

(iii)

Let H be the centralizer of A in G. Then H is a normal subgroup of G containing A and it will be sufficient to show that H = A. Let X/A be the center of G/A and suppose H > A. Then we see from (ii) that X/A $\cap$ H/A > 1 because H/A $\vartriangleleft$ G/A, and we deduce that there is a nontrivial cyclic subgroup Y/A contained in X/A $\cap$ H/A. Since Y/A $\leqslant$ X/A we see that Y/A $\vartriangleleft$ G and we deduce that Y $\vartriangleleft$ G. Also A is contained in the center of Y and Y/A is cyclic, hence Y is abelian. This contradicts the fact that A is a maximal normal abelian subgroup of G and the result follows

2.

(i)

The number of Sylow 5-subgroups of G divides 36 and is congruent to 1 modulo 5, which means that this number must be 1,6 or 36. It cannot be 1, for then G would have a normal Sylow 5-subgroup, which contradicts the fact that G is simple. Nor can G have 6 Sylow 5-subgroups, for then G would be isomorphic to a subgroup of A₆ because G is simple. This would mean that A₆ has a subgroup of index 2, and since subgroups of index 2 are always normal, this would contradict the fact that A₆ is simple. We conclude that G has 36 Sylow 5-subgroups.

(ii)

Let N be the normalizer of a Sylow 3-subgroup. Then the number of Sylow 3-subgroups is [G : N]. Also the number of Sylow 3-subgroups divides 20 and is congruent to 1 modulo 3, so this number is 1,4 or 10. It cannot be 1 because that would mean G has a normal Sylow 3-subgroup, which would contradict the fact that G is simple. Nor can it be 4, for then G would be isomorphic to a subgroup of A₄ because G is simple, which is clearly impossible. Therefore the number of Sylow 3-subgroups is 10. We deduce that [G : N] = 10 and hence N has order 18.

(iii)

A Sylow 3-subgroup of a group of order 18 has order 9. This means the subgroup has index 2, hence the subgroup is normal because in any group, a subgroup of index 2 is normal.

(d)

Let C be the centralizer in G of A $\cap$ B and suppose A $\cap$ B is not 1. Then C contains A, B and A $\cap$ B is a normal subgroup in C, so C cannot be the whole of G. Therefore the order of C is a multiple of 9 and divides 180, and is neither 9 nor 180. Let d be the index of C in G, so | C| = | G|/d. Then G is isomorphic to a subgroup of A_d because G is simple. Since the order of A_d is less than 180 if d$\le$5, we see that d$\ge$6 and consequently C has order 18. From part (iii), the Sylow 3-subgroup of C is normal in C and therefore C has exactly one subgroup of order 9. We now have a contradiction because C has two subgroups of order 9, namely A and B. We conclude that A $\cap$ B = 1.

(e)

We count the elements in G. Since two Sylow 5-subgroups intersect in 1 and there are 36 Sylow 5-subgroups by (i), we see that G has 36*4 = 144 elements of order 5. Also two Sylow 3-subgroups intersect in 1 by (iv) and G has 10 Sylow 3-subgroups by (ii). Therefore G has 8*10 = 80 elements of order 3 or 9. We conclude that G has at least 144 + 80 = 224 elements, which contradicts the fact that G has only 180 elements. It follows that no such G can exist and thus there is no simple group of order 180.

3.

Since M is a cyclic R-module, we know that M $\cong$ R/Rs for some s $\in$ R. By the uniqueness part of the fundamental structure theorem for finitely generated modules over a PID, we cannot write R = A $\oplus$ B where A, B are nonzero R-modules. Therefore s$\ne$ 0. Since sM = 0, we may take r = s.

Since rM = 0, we see that rR $\subseteq$ sR and hence s divides r. Suppose there do not exist distinct primes p, q dividing r. Then the same is true for s because s divides r, and we deduce that s is a prime power, say p^e for some prime p. From the uniqueness statement in the fundamental structure theorem for finitely generated modules over a PID, we cannot write R/Rp^e = A $\oplus$ B where A, B are nonzero R-modules and we have a contradiction. This finishes the proof.

4.

(i)

Choose integers r, s such that qr + 2s = 1. Then in $\mathbb {Z}$[[X]]/(X - 2) we have qr = 1 - sX mod (X - 2). Since 1 - sX is invertible in $\mathbb {Z}$[[X]] (with inverse 1 + sX + s²X² + ...), it follows that q is invertible in $\mathbb {Z}$[[X]]/(X - 2).

(ii)

The general element of R is $\sum$a_iXⁱ mod (X - 2), where a_i $\in$ $\mathbb {Z} _{(2)}^{}$ for all i. Now each a_i is of the form p/q where p, q $\in$ $\mathbb {Z}$ and q is odd. Using (i), we may now write a_i = b_i mod (X - 2) where b_i $\in$ $\mathbb {Z}$[[X]], and then we may write the general element of R in the form $\sum$b_iXⁱ mod (X - 2). This proves that $\pi$$\theta$ is surjective. We now determine the kernel of $\pi$$\theta$ . Obviously (X - 2) $\subseteq$ ker $\pi$$\theta$ . Conversely suppose f $\in$ ker $\pi$$\theta$ . Then we may write f = (X - 2)g where g $\in$ $\mathbb {Z} _{(2)}^{}$[[X]]. We want to show that g $\in$ $\mathbb {Z}$[[X]]. Write g = $\sum$g_iXⁱ where g_i $\in$ $\mathbb {Z} _{(2)}^{}$ . Then the coefficient of Xⁿ in g(X - 2) is g_{n - 1} - 2g_n for n > 0, and the constant coefficient is -2g₀. By induction on n we see that 2g_n $\in$ $\mathbb {Z}$ and since g_n $\in$ $\mathbb {Z} _{(2)}^{}$ , we conclude that g_n $\in$ $\mathbb {Z}$ for all n. This proves that ker $\pi$$\theta$ = (X - 2) and it now follows from the fundamental isomorphism theorem that R $\cong$ $\mathbb {Z}$[[X]]/(X - 2).

(iii)

By considering the homomorphism $\mathbb {Z}$[X]$\to$$\mathbb {Z}$ determined by sending X to 2, we see that $\mathbb {Z}$[X]/(X - 2) $\cong$ $\mathbb {Z}$ . Since 3 is not invertible in $\mathbb {Z}$ , we see that 3 is not invertible in $\mathbb {Z}$[X]/(X - 2). But 3 is invertible in R and the result follows.

5.

(i)

Obviously K($\alpha^{p}_{}$) $\subseteq$ K($\alpha$). Also $\alpha$ is separable over K($\alpha^{p}_{}$) and satisfies the polynomial X^p - $\alpha^{p}_{}$ . Since $\alpha$ is the only root of X^p - $\alpha^{p}_{}$ , it follows that $\alpha$ $\in$ K($\alpha^{p}_{}$) and hence K($\alpha$) = K($\alpha^{p}_{}$). Now we consider the minimum polynomial of $\beta$ over K. This has degree p because [K($\beta$) : K] = p, and must be a polynomial in X^p because $\beta$ is not separable over K. Thus the minimum polynomial must be of the form X^p - b for some b $\in$ K and it follows that $\beta^{p}_{}$ = b $\in$ K.

(ii)

Since we are in characteristic p, we have ($\alpha$ + $\beta$)^p = $\alpha^{p}_{}$ + $\beta^{p}_{}$ . But $\beta^{p}_{}$ $\in$ K by (i), hence $\alpha^{p}_{}$ $\in$ K($\alpha$ + $\beta$). Therefore K($\alpha^{p}_{}$) $\subseteq$ K($\alpha$ + $\beta$) and it now follows from (i) that K($\alpha$) $\subseteq$ K($\alpha$ + $\beta$) as required.

(iii)

Since K($\alpha$ + $\beta$) $\supseteq$ K($\alpha$) by (ii), we have [K($\alpha$ + $\beta$) : K] = [K($\alpha$ + $\beta$) : K($\alpha$)][K($\alpha$) : K] and since [K($\alpha$) : K] = d, it remains to prove that [K($\alpha$ + $\beta$) : K($\alpha$)] = p. Now ($\alpha$ + $\beta$)^p = $\alpha^{p}_{}$ + $\beta^{p}_{}$ $\in$ K($\alpha$) which shows that [K($\alpha$ + $\beta$) : K($\alpha$)] = p or 1, because we are in characteristic p. It remains to prove that [K($\alpha$ + $\beta$) : K($\alpha$)] $\neq$ 1, or equivalently that $\beta$$\notin$K($\alpha$). But $\alpha$ is separable over K, hence every element of K($\alpha$) is separable over K which shows that $\beta$$\notin$K($\alpha$) because $\beta$ is not separable over K. This completes the proof.

6.

(i)

It will be sufficient to prove that X^p - t is irreducible in $\mathbb {C}$[t, X], or equivalently that t - X^p is irreducible in $\mathbb {C}$[X, t]. Let R = $\mathbb {C}$[X] and let F = $\mathbb {C}$(X), the field of fractions of R. Then t - X^p is a monic polynomial in R[t] and is irreducible in F[t], hence it is irreducible in R[t] = $\mathbb {C}$[X, t] and the result follows.

(ii)

Let y be one of the roots of X^p - t in L. Since X^p - t is irreducible in K[X], we see that [K(y) : K] = p. Also the roots of X^p - t are e^2n$\pi$i/py where n = 0,..., p - 1, and since e^2n$\pi$i/p $\in$ $\mathbb {C}$ for all n, we deduce that all the roots of X^p - t are in K(y). It follows that K(y) = L and we conclude that [L : K] = p. Therefore the Galois group of L over K has order p (note that L/K is a separable extension because we are in characteristic zero). Since groups of order p are cyclic, we conclude that the Galois group of L over K is cyclic of order p and hence isomorphic to $\mathbb {Z}$/p$\mathbb {Z}$ .