External and Internal Direct Product -- Abstract Algebra Examples 13

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[M.athematech]

14:58 you proved both set inclusion directions needed to prove the reverse implication, but you forgot to prove the forward implication. And you make the same mistake on the next problem at 18:30.

[schweinmachtbree]

This problem set is quite strangely written - the penultimate two exercises would be much better phrased as "... prove that <m, n> = <d> where d = gcd(m, n)" and "... prove that <m> ∩ <n> = <l> where l = lcm(m, n)", and the first exercise could be clarified by saying "Prove for all _n_ that ..." (which is what was being asked) since it wasn't clear to Justin.

Justin's method for _n_ = 1 works great (a good way of showing that things are not isomorphic is by looking at invariants, i.e. things that don't change under isomorphism), but sadly it doesn't generalise to other values of _n_ . To solve it in general one uses the equivalence between external and internal direct products: if D_4n were isomorphic to the external direct product D_2n × *Z* _2 then D_4n would be an internal direct product of a subgroup isomorphic to D_2n and a subgroup isomorphic to *Z* _2. These subgroups have to be normal by the definition of internal direct product, and subgroups isomorphic to *Z* _2 correspond to elements of order 2: one can easily check that <t> is not normal for any reflection t, so the only remaining subgroup of order 2 is <r^2n>, which is normal - indeed it is the center of D_4n, and therefore it will also commute with any subgroup, which is another requirement for internal direct products.

It is not too hard to prove that for a divisor k of m, the dihedral group D_m has subgroups isomorphic to D_k; namely it has m/k such subgroups. In our case we have m = 4n and k = 2n, so there are 2 subgroups of D_4n isomorphic to D_2n, call them H and K - these correspond to two ways of inscribing a regular 2n-gon inside a regular 4n-gon (either with the vertices of the 2n-gon lining up with vertices of the 4n-gon, or with the vertices of the 2n-gon lining up with edges of the 4n-gon). Since these subgroups have index 2 they are necessarily normal, so in order to show that D_4n is _not_ isomorphic to D_2n × *Z* _2 we need to show that the final remaining criterion of internal direct products fails: that <r^2n> does not intersect trivially with H or with K. H and K are given by <r^2, s> = <1, r^2, r^4, ..., r^(4n-2), s, sr^2, sr^4, ..., sr^(4n-2)> and <r^2, sr> = <1, r^2, r^4, ..., r^(4n-2), sr, sr^3, sr^5, ..., sr^(4n-1)>, and we can see that they have non-trivial intersection with <r^2n> (namely <r^2, s> ∩ <r^2n> = {1, r^2n} = <r^2n> and likewise <r^2, sr> ∩ <r^2n> = <r^2n>).

[richardschreier]

The last problem (18:32) asks for *internal* direct product representations of U₁₆₅ ≅ U₃ × U₅ × U₁₁. Since U₃, U₅ and U₁₁ are cyclic groups of order 2, 4 and 10, respectively, we are looking for cyclic subgroups N₁, N₂ and N₃ of U₁₆₅ such that |N₁| = 2, |N₂| = 4, |N₃| = 10, and N₁N₂N₃ = U₁₆₅. The latter condition requires N₁ ⋂ N₂ = N₁ ⋂ N₃ = N₂ ⋂ N₃ = {1}, which can be used to eliminate some candidates before computing N₁N₂N₃. A computer search yields many solutions, including (N₁, N₂, N₃) = (<56>, <23>, <19>) and (N₁, N₂, N₃) = (<164>, < 142>, <151>). Note that of these 3-subgroup solutions also yields three 2-subgroup solutions to the problem, e.g. U₁₆₅ = <56, 23><19> = <56> <23, 19> = <56, 19> <23>, so there are many solutions to choose from.

[davidemasi__]

9:05 do you actually need to show it is well defined? I mean, since the domain is not a set of equivalence classes (x, y)=(m, n) means x=m and y=n, so there's no such thing as "same element with different names", right?

[M.athematech]

The way the first problem is worded one should show D4n is not isomorphic to D2n x Z2 for any n, but you only show it for n = 1.

[finnr]

Why is he using <5> in the last example, whereas 5 isnt even in U_165 since 5 isn't coprime to 165?

[M.athematech]

Proof of being well-defined at 9:09 is pointless nonsense, you aren't defining a function using possibly differing representatives of an equivalence class.

[schweinmachtbree]

A couple mistakes: 1:57 is badly phrased - what you mean is that if two groups are isomorphic they are either both abelian or both non-abelian, or more generally if phi: G_1 → G_2 is an isomorphism then x and y commute in G_1 if and only if phi(x) and phi(y) commute in G_2. however saying "we must preserve commutativity" makes it sound like we have a say in the matter, which we don't because G_1 = D_4n and G_2 = D_2n × *Z* _2 are given - better would be "commutativity must be preserved"/"commutativity must be the same in both groups" without the 'we'.
At 2:05 it is clearly not true that e and r^2 are the only elements that commute in D_4; for example any r^i commutes with any r^j, and of course any element commutes with itself. what you meant, as you say a bit later, is that e and r^2 are the only elements of D_4 that commute with _everything_, i.e. that {e, r^2} is the center of D_4, but you don't need to argue with centers to show that D_4 and D_2 × *Z* _2 are non-isomorphic because you can just note that D_4 is not abelian but D_2 × *Z* _2 is. (If you did want to argue with centers then you would have to first show how centers interact with direct products; you compared Z(D_4) and Z(D_2) instead of Z(D_4) and Z(D_2 × *Z* _2).)
Lastly there is a typo at 16:35 - it should be "0 ≤ r < l".

[iGeen]

Commutativity and associativity of 'products' has not been proved.... (though, it is simple)