Oops!  The second midterm turned out to be harder than I intended.
Also I graded a bit more strictly this time (although equally so for
everyone).  Thus the grades were a lot lower than for the first
midterm: the highest score was 40 and the median was 23.5.  It doesn't
really matter, since the final grades will be curved at the end of the
course.  However, if you want to improve, I think it is important to
go over each question, make sure you understand how to do it, and fill
in any gaps in your understanding that the midterm may have exposed.
I am also happy to help during office hours.

Here is the score distribution:

Score range	Number of tests

36-40		2
31-35		6
26-30		9
21-25		8
16-20		8
11-15		3
6-10		1
0-5		1

Here are comments on some points that were particularly troublesome:

1a) Hardly anyone got this completely right.  Please see the solution
sheet for one correct answer.

2a) Many people said something to the effect that if H is a normal
subgroup of G then ghg'=h for all g in G and h in H.  (Here g' denotes
the inverse of g.)  However this is not true.  All you know is that
ghg' is some element of H, which may be different from h.  An example
where this happens is G=S_3 and H=A_3.  (Try it.)  A simpler example
is when G is any nonabelian group and H=G.  If ghg'=h for all g in G
and h in H, then H is a subgroup of the center of G, which is a much
stronger condition than being normal, i.e. it implies normality but
doesn't follow from it.

4a) It is important here to do everything in Z_11.  Also, since we are
dividing by something of degree 1, the remainder must have degree less
than one, i.e. it must be a constant.  Here is the complete long
division:

	3x^2 + 0x   + 10    remainder 9
	__________________________________
4x - 1 | x^3 + 8x^2 + 7x - 1
	 x^3 - 3x^2
	 ----------      
	       0x^2  +7x - 1
		      7x - 10
	            -----------
			   9 

5a) Hardly anyone got this completely right.  Please see the solution
sheet for one correct answer.  By the way, this quotient is isomorphic
to Z x Z_2.  However it takes a bit of work to prove this.  Here is an
outline of one proof.  First let's prove something a bit easier: Z x Z
/ <(0,2)> is isomorphic to Z x Z_2.  One defines an isomorphism by
sending the coset of (a,b) to (a,b mod 2).  Now one can reduce what we
want to this by changing coordinates.  Namely by linear algebra there
is an automorphism of Z x Z sending (1,0) to (1,1) and (0,1) to (1,2).
To see that this works you have to check that the inverse of the
corresponding matrix has integer entries, which holds because the
determinant is 1 or -1.  So every element of Z x Z can be uniquely
written as a(1,1) + b(1,2) where a and b are integers, and we can
define an isomorphism from Z x Z / <(2,4)> to Z x Z_2 by sending an
element of Z x Z to (a,b mod 2).  A generalization of this argument
shows that if (x,y) is a nonzero element of Z x Z, and if gcd(x,y) =
d, then Z x Z / <(x,y)> is isomorphic to Z x Z_d.