## 330 Problem Set 4 Comments

13.45 Most everyone who tried 13.45 did fine.  By homomorphism it’s a subgroup, since the target is finite, it has finite order, and by Lagrange’s theorem the order of the subgroup divides the whole order.

13.46 7.6 says that since G is generated by {a_i}, every element in G is a product of powers of elements in the generating set.  so, both \phi(g) and \mu(g) can be broken up into what they do on the generators, since they do the same thing, they must be equal.  There are details, but those are the ideas.

13.48 The kernel of sgn is the even permutations, i.e. A_n, all of those that get mapped to +1, the identity in  the multiplicative group {±1}.  A statement  of the answer is not sufficient.  You need the reason given in the first sentence.  This is similar to 13.3, where the additive group {0,1} is used instead.

13.49 is rather easy also - apply homomorphism property twice.  Just make sure since it’s so short that you’re explicit about what is used where.

13.50 Show that \phi[G] is abelian by showing that for all \phi(x) and \phi(y) in \phi[G], \phi(x)\phi(y) = \phi(y)\phi(x).  This is the same as \phi(x)\phi(y)\phi(x)^{-1}\phi(y)^{-1}  = e_H.  By homomorphism property the left is the same as \phi(xyx^{-1}y^{-1}), which says that xyx^{-1}y^{-1} is in the kernel because it goes to the identity.  The steps are reversible.

13.51 unattempted

13.52 Is the other half of what we did in class for 13.15, and is the same.

For all 13.53-55 please remember that “necessary and sufficient” is different language for “if and only if”.  Very few proved both directions.  It is incomplete (2)  if you did not.  Some proved both directions “by accident”, and didn’t explicitly state what they were doing.  This is 3.5.

13.53 kh=hk is the condition.  In one direction the key is \phi(1,1)*\phi(1,1)=\phi(2,2).  This gives hkhk=h^2k^2, and canceling left and right gives kh = hk, as desired.  Or even better \phi(1,0)\phi(0,1) = \phi(1,1) = \phi(0,1)\phi(1,0).  The other direction is more natural.

13.54 G is abelian is the condition, using the same reasoning.

13.55 Please do not use |h| for the order of an element h.  Either use “the order of element h” or |<h>|, which is the order of the cyclic subgroup generated by h, which is the definition of the order of the element.  |h| rightly looks like absolute value.  Please.  Here the condition is either h^n = e or the order of h divides n.   They say the same thing.  In one direction you need to prove that if \phi is a homomorphism then h^n=e.  Here’s a proof e=\phi(0) = \phi(1 + (n-1)) = \phi(1)\phi(n-1)=h^1h^{n-1}=h^n.

14.26 I thought that part of this question is to show that T is a subgroup, as it is not demonstrated anywhere else.  I was wrong.  This is apparently 11.39.  So, the first part is to show that T is normal, which is trivial because G is abelian.  The only remaining part is to show that G/T is torsion free.  That is moderately interesting.  Suppose (gT)^n = T (the  identity in T).  We seek to show that gT = T, i.e. that g is in T.  T=(gT)^n = g^nT (given how we multiply cosets.  So, we have that g^n is in T.  But that means that g^n is of finite order, say q, so g^(n^q) = e, i.e. g is of finite order so in T, as requested.

14.27 Show that conjugacy  as a relation is reflexive symmetric and transitive.   i_e takes H to itself.  if i_g takes H to K, then i_g^{-1}  takes K to H.  And if I_g takes H to K and i_j takes K to  L, then I_jg takes H to L, so we have transitivity.    There are details, but these are the ideas.

14.28 N is normal iff the only subgroup conjugate to it is itself.  This requires proving, but yes, the key step is condition #2 of normal.

14.31 Does not say how many sets, so may not assume how many.  Must prove for an arbitrary  number (possibly infinite).  In the end the proof for 2 sets is about the same, but you must acknowledge that  it isn’t necessarily two.   For those who didn’t, 7.4 is a good place to look for how to do this.

14.32 not attempted, but it is the intersection of all normal subgroups containing S.

14.33 We want to show that aCbC = bCaC, i.e. that abC = baC.  So, we start with abc and try to show that it equals some bac’.  ba(a^{-1}b^{-1}ab)=ab, and so ba(a^{-1}b^{-1}ab)c = abc and since all commutators are in c, (a^{-1}b^{-1}ab)c = c’ is in C.  The other inclusion can be shown by switching letters.  Be careful, C is generated by commutators, but every element in C is not a commutator itself.  Notice I did not need that.

14.34 In a video I suggested doing this by contrapositive:
If H is abnormal (no, people don’t say that) then G  has more than one subgroup of a given order.
If H is abnormal gHg^{-1} ≠  H for some g, but gHg^{-1} is a subgroup the image of the inner automorphism.  So, there is more than one subgroup.

14.35 Showing H and N = M is normal in H is a two step process, because H is a subgroup, hnh{-1} is inside H, and because N is normal hnh{-1} is inside N, so it is in both, and hence in M.  I was thinking for a counterexample of G = D_4, N = the rectangle group, and H = {0,H} (the second H is the horizontal reflection).  We saw in a video that H is not normal in D_4.  An even easier example is in the smallest nonabelian group, S_3.  Let N = S_3 itself (sneaky), and H = {i,(12)} which is not a normal subgroup.