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Abstract Algebra

  1. Mar 22, 2005 #1
    Question 1
    Determine the quotient group [tex](\mathbb{Z}_2\times\mathbb{Z}_4)/\langle(1,2)\rangle[/tex]

    [tex]\langle(1,2)\rangle[/tex] is a cyclic subgroup [tex]H[/tex] of [tex]\mathbb{Z}_2\times\mathbb{Z}_4[/tex] generated by [tex](1,2)[/tex]. Thus


    Since [tex]\mathbb{Z}_2\times\mathbb{Z}_4[/tex] has 2.4 = 8 elements, and [tex]H[/tex] has 2 elements, all cosets of [tex]H[/tex] must have 2 elements, and [tex](\mathbb{Z}_2\times\mathbb{Z}_4)/H[/tex] must have order 4.

    Possible abelian groups of order 4 are


    But I dont know how to work out which one is isomorphic to [tex](\mathbb{Z}_2\times\mathbb{Z}_4)/\langle(1,2)\rangle[/tex]
  2. jcsd
  3. Mar 22, 2005 #2
    Which of [tex] \mathbb{Z}_2 \times \mathbb{Z}_2[/tex], [tex] \mathbb{Z}_1 \times \mathbb{Z}_4[/tex], and [tex](\mathbb{Z}_2 \times \mathbb{Z}_4)/H[/tex] are cyclic?
  4. Mar 22, 2005 #3
    I think [tex]\mathbb{Z}\times\mathbb{Z}_4[/tex] is cyclic because it is generated by [tex]a^n[/tex] for all [tex]n=0,1,\dots[/tex] where [tex]a=(1,2)[/tex] and [tex]n = 2[/tex]
  5. Mar 22, 2005 #4
    Well, as I posted, you really mean [itex]\mathbb{Z}_1 \times \mathbb{Z}_4[/itex] (or equivalently just [itex]\mathbb{Z}_4[/itex]), since [itex]\mathbb{Z} \times \mathbb{Z}_4[/itex] doesn't have order 4 at all (not even finite order!).

    But it is indeed cyclic (generated by [itex](0, 1)[/itex] or just [itex]1[/itex] if you choose the [itex]\mathbb{Z}_4[/itex] variety~).

    Is [itex](\mathbb{Z}_2 \times \mathbb{Z}_4)/H[/itex] cyclic?
    Last edited: Mar 22, 2005
  6. Mar 22, 2005 #5
    Yes, it is cyclic. Isn't there a theorem that says a quotient group of cyclic group is cyclic?

    So wouldn't [tex]\mathbb{Z}\times\mathbb{Z}_4[/tex] be the answer? Or am I missing something?
  7. Mar 22, 2005 #6
    I don't remember such a theorem. Let's just write down the elements of [itex](\mathbb{Z}_2 \times \mathbb{Z}_4)/H[/tex]. This set is just

    [tex] \{ (1, 0) + H, \; (1, 1) + H, \; (0, 1) + H, \; (0, 0) + H \}[/tex].

    Does it have a generator? Remember the group is abelian and [itex] H + H = H[/itex].

    Edit: Actually, there may be. I really don't remember! :smile: Anyways, in this case the group in question is indeed cyclic (and thus isomorphic to [itex]\mathbb{Z}_4[/itex]).
    Last edited: Mar 22, 2005
  8. Mar 22, 2005 #7
    [tex]\mathbb{Z}_2\times\mathbb{Z}_4 = \{(0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2),(1,3)\}[/tex]

    But this can be collapsed into

    [tex]\mathbb{Z}_4 = \{(0,0),(0,1),(0,2),(0,3)\} = \mathbb{Z}_2\times\mathbb{Z}_2[/tex]

    So then

    [tex](\mathbb{Z}_2\times\mathbb{Z}_4)/H = \{(0,0)+H,(0,1)+H,(0,2)+H,(0,3)+H\}[/tex]

    where [tex]H=\langle(1,2)\rangle[/tex]

    and its generator is [tex](1,2)[/tex]
    Last edited: Mar 22, 2005
  9. Mar 22, 2005 #8
    I think the theorem you're thinking of is that if [itex]G[/itex] is cyclic then every quotient group [itex]G/H[/itex] is cyclic. In our case we can't use that (since [itex] H[/itex] is cyclic, not [itex]\mathbb{Z}_2 \times \mathbb{Z}_4[/itex]). There's no theorem the way we need it :frown:
    Last edited: Mar 22, 2005
  10. Mar 22, 2005 #9
    I wouldn't include the line saying "can be collapsed into" (what does that mean?). And I think in your second TeX line (the one I think you should remove, because the first equality isn't true~) you meant [itex] = \mathbb{Z}_1 \times \mathbb{Z}_4[/itex], not [itex] = \mathbb{Z}_2 \times \mathbb{Z}_2[/itex]!

    The rest looks good though~
    Last edited: Mar 22, 2005
  11. Mar 22, 2005 #10
    Ok, I fixed it up a bit.

    Let me get this straight. The quotient group under investigation is

    Here [tex]\langle(1,2)\rangle[/tex] is the cyclic subgroup [tex]H[/tex] of [tex]\mathbb{Z}_2\times\mathbb{Z}_4[/tex] generated by [tex](1,2)[/tex]. Thus

    [tex]H = \{(0,0),(1,2)\}[/tex]

    Since [tex]\mathbb{Z}_2\times\math{Z}_4[/tex] has 8 elements and [tex]H[/tex] has 2 elements, all cosets of [tex]H[/tex] must have 4 elements, and [tex](\mathbb{Z}_2\times\mathbb{Z}_4)/H[/tex].

    In additive notation, the cosets are

    [tex]H = \{(0,0)+H,(0,1)+H,(0,2)+H,(0,3)+H\}[/tex]

    Since we can compute by choosing representatives [tex](0,0),(0,1),(0,2),(0,3)[/tex] it is clear that [tex](\mathbb{Z}_2\times\mathbb{Z}_4)/H[/tex] is isomorphic to [tex]\mathbb{Z}_4[/tex]

    Note that this is what we would expect, since in a factor modulo [tex]H[/tex], everything in [tex]H[/tex] becomes the identity element; that is we are essentially setting everthing in [tex]H[/tex] to zero. Thus the whole factor [tex]\mathbb{Z}_2[/tex] of [tex]\mathbb{Z}_2\times\mathbb{Z}_4[/tex] is collapsed, leaving just the second factor [tex]\mathbb{Z}_4[/tex].
  12. Mar 22, 2005 #11
    looks good :smile:
  13. Mar 22, 2005 #12
    Question 2

    Show that ifa group [tex]G[/tex] has exactly one subgroup [tex]H[/tex] of given order, then this subgroup must be normal.

    I figured that since every group has at least two subgroups: the whole group itself (the improper subgroup) [tex]G[/tex] and the trivial subgroup [tex]\{e\}[/tex].

    If a group [tex]G[/tex] has exactly one subgroup then [tex]G = \{e\}[/tex] and the group in question is actually the trivial group. This is the only way that I can see that a group can have exactly one subgroup - if it is the trivial group.

    So if [tex]G[/tex] is the trivial group, then it consists of only one element, namely the identity. The group consisting of one element, [tex]e[/tex] is commutative. ie [tex]e*a=a*e[/tex] where [tex]a \in G[/tex]. Obviously [tex]a=e[/tex] if [tex]G[/tex] has only one element.

    Hence [tex]G[/tex] is abelian. And all subgroups of abelian groups are normal.


    A subgroup is normal if its left and right cosets coincide. That is, [tex]gH=Hg \, \forall \, g \in G[/tex]. Since [tex]g=e[/tex] then we have [tex]eH=He[/tex] which is trivial because [tex]G[/tex] is abelian.

    Therefore the subgroup [tex]H[/tex] of [tex]G[/tex] is normal.
  14. Mar 22, 2005 #13
    You are misinterpreting the question (at least the way I read it). It is not saying that [itex]G[/itex] has only one subgroup. It says that [itex]G[/itex] has only one subgroup [itex]H[/itex] of a particular order, say [itex]n[/itex], ie. if [itex] J \leq G[/itex] and [itex] |J| = n[/itex] then [itex]J = H[/itex].
  15. Mar 22, 2005 #14
    Ok, lets assume that you read the question correctly (which is likely). Then I have to show that the subgroup is normal given that it is the only subgroup with a particular order [tex]n[/tex].

    This means that [tex]G[/tex] has only one subgroup with [tex]n[/tex] elements, all other subgroups have some other order.

    For a subgroup of a group to be normal, then the following are equivalent
    [tex]ghg^{-1} \in H \, \forall \, g\in G \mbox{and} h\in H[/tex]
    [tex]gHg^{-1} = H \, \forall \, g\in G[/tex]
    [tex]gH = Hg \, \forall g\in G[/tex]

    But I don't know how to start. That is, none of the conditions for a subgroup to be normal, involve the order of the subgroup.

    Any hints?
    Last edited: Mar 22, 2005
  16. Mar 23, 2005 #15
    N normal <==> NaNb = Nab for a, b not in subgroup N
  17. Mar 24, 2005 #16
    Sorry I haven't replied in a while.

    Here's a hint:

    Recall that [itex]H[/itex] normal in [itex]G[/itex] if and only if [itex]gHg^{-1} = H \mbox{ for every } g \ \mbox{in} \ G[/itex]. Assume [itex]H[/itex] isn't normal. Then there is some [itex]g[/itex] in [itex]G[/itex] such that [itex]gHg^{-1} \neq H[/itex]. For this [itex]g[/itex], consider the set [itex]gHg^{-1}[/itex] (Hint: Try to prove that it is a subgroup of [itex]G[/itex] of the same order as [itex]H[/itex], to get a contradiction).
    Last edited: Mar 24, 2005
  18. Mar 24, 2005 #17
    Figured I'd put this back at the top in case you still need it, since you're online right now~
  19. Mar 24, 2005 #18
    The condition for [tex]H[/tex] being normal is

    [tex]ghg^{-1} \in H\quad \forall g \in G,\, \forall h \in H[/tex]

    Assume that [tex]H[/tex] isn't normal (proof by contradiction)

    Then [tex]\exists \, k \in G[/tex] such that

    [tex]khk^{-1} \notin H \quad \forall h \in H[/tex]

    Now consider the set

    [tex]A = \{khk^{-1} | \, \forall h \in H\}[/tex]

    We now show that this is a subgroup of [tex]G[/tex] and that it has the same order as [tex]H[/tex].

    For [tex]h_1,h_2 \in H, \, \exists\, kh_1k^{-1},\, kh_2k^{-1} \in A[/tex] by definition. Now,

    [tex](kh_1k^{-1})(kh_2k^{-1}) = (kh_1)(k^{-1}k)(h_2k^{-1})[/tex]
    [tex]= kh_1(e)h_2k^{-1}[/tex]
    [tex]= k(h_1h_2)k^{-1}[/tex]

    But [tex]h_1,h_2 \in H[/tex] as [tex]H[/tex] is a subgroup by assumption.

    Hence [tex]kh_1,h_2k^{-1} \in A[/tex] which implies that [tex]A[/tex] is a subgroup of [tex]G[/tex]

    Now so far we have been using the fact that [tex]H[/tex] is not normal. And from this we have got a subgroup [tex]A[/tex]. But it is obvious that for each [tex]h \in H[/tex] we can make an [tex]a \in A[/tex] by

    [tex]kh_1k^{-1} = kh_2k^{-1} \Leftrightarrow h_1 = h_2[/tex]

    the left and right cancellation laws. Hence [tex]H[/tex] and [tex]A[/tex] have the same order. Alternatively I think I could have proven this by setting a bijection. Anyway, [tex]H[/tex] and [tex]A[/tex] having the same order is a contradiction since [tex]H[/tex] is the only subgroup with a given order. Hence [tex]H[/tex] MUST be normal.
    Last edited: Mar 24, 2005
  20. Mar 24, 2005 #19
    Everything looks great, except for one detail:

    You cannot assume [itex] \exists k \in G[/itex] such that

    [tex]khk^{-1} \notin H \; \forall h \in H[/tex].

    This is too strong.

    All you can assume is

    [itex] \exists k \in G[/itex] such that

    [tex]\{ khk^{-1} | h \in H \} \neq \{ h \in H\}[/tex]

    there is a subtle difference, but it does not affect the rest of the proof. :smile:

    Also, you must show that [itex] e \in A[/itex], but this is trivial. Also, that [itex] x \in A \Longrightarrow x^{-1} \in A[/itex] (also easy).

    You are quite correct that the trivial bijection (and, in fact, isomorphism)

    [tex] f : H \longrightarrow A[/tex]
    [tex] \forall \ h \in H, \ f(h) = khk^{-1}[/tex]

    would also have worked nicely :smile:
    Last edited: Mar 25, 2005
  21. Mar 25, 2005 #20
    Thanks Data for your input. I'd have to say, you have a knack for mentoring.

    Here is another question I have been working on.

    Let [tex]\textsc{Q}[/tex] be the subgroup of [tex]GL(2,\mathbb{C})[/tex] generated by

    [tex]\left(\begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array}\right) \quad \mbox{and} \quad \left(\begin{array}{cc} 0 & i \\ i & 0 \\ \end{array}\right)[/tex]

    1. Show that [tex]\textsc{Q}[/tex] is a nonabelian group of order 8.
    2. Is [tex]\textsc{Q}[/tex] isomorphic to the dihedral group [tex]\mathcal{D}_4[/tex]?
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