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I want to show that $$U(3)×U(5)\cong U(15)$$ Would I simply have to find an isomorphic map that maps the two groups, or is there a clever way to approach this? I have been trying to find an isomorphic map but have had no luck. Does anyone know of an isomorphic map that maps these two groups?

$U(3) \times U(5) = \left \{ (1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)\right \} $ and $U(15) = \left \{ 1,2,4,7,8,11,13,14\right \}$

MomoTheSir
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2 Answers2

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This is an application of the chinese remainder theorem: https://en.wikipedia.org/wiki/Chinese_remainder_theorem

If you can't come up with the map yourself, look at the map in the proof of CRT and try to prove it is an isomorphism yourself.

GiantTortoise1729
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The elements of order 2 in $U(3)×U(5)$ are $(2,1)$, $(1,4)$ and $(2,4)$. These should map to elements of order 2 in $U(15)$: 4, 11 and 14. We can have for example $$(1,1)\mapsto1$$ $$(2,1)\mapsto11$$ $$(1,4)\mapsto4$$ $$(2,4)\mapsto14$$ and these elements form a Klein-four subgroup of both groups. The other four elements have order 4, and we can have for example $$(1,2)\mapsto2$$ $$(2,2)\mapsto7$$ $$(1,3)\mapsto8$$ $$(2,3)\mapsto13$$ which gives an isomorphism between $U(3)×U(5)$ and $U(15)$ as $$\begin{array}{c|cccc} (a,b)&1&2&4&3\\ \hline 1&1&2&4&8\\ 2&11&7&14&13 \end{array}$$

  • Multiplication by 2 in $U(3)$ corresponds to multiplication by 11 in $U(15)$.
  • Multiplication by 2 in $U(5)$ corresponds to multiplication by 2 in $U(15)$.
Parcly Taxel
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  • This deserves a positive up-vote. – Mikasa Nov 07 '16 at 05:17
  • Thank you for the well detailed response, but I do not understand the last two bullets. Additionally, what map function did you use to map elements of $U(3) \times U(5)$ to $U(15)$? – MomoTheSir Nov 07 '16 at 16:00
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    @MomoTheSir The mapping is given explicitly in the answer; I just assigned mappings based on the order of the elements. The last two bullets tell you that (for example) if $(a,b)$ is in $U(3)×U(5)$ and it maps to $c$ in $U(15)$, then $(2a,b)$ maps to $11c$. – Parcly Taxel Nov 07 '16 at 16:06
  • Because there are 3 elements of order 2 in both groups can I just map any of the 3 elements of order 2 in $U(3) \times U(5)$ to any of the 3 elements of order 2 in $U(15)$? – MomoTheSir Nov 07 '16 at 16:16
  • @MomoTheSir The mapping is not arbitrary. All order-4 elements $a$ in $U(15)$ and $U(3)×U(5)$ have $a^2=4$, so $(1,4)\mapsto4$ is forced. There is a choice in assigning $(2,1)$ and $(2,4)$ to 11 and 14; after this, assigning any remaining element to $(1,2)$ fixes everything. – Parcly Taxel Nov 07 '16 at 16:29
  • How did you determine that $(2,1) \rightarrow 11$ and $(2,4) \rightarrow 14$? Does it matter if I assign $(2,1) \rightarrow 14$ and $(2,4) \rightarrow 11$ instead? – MomoTheSir Nov 07 '16 at 17:05
  • @MomoTheSir Doesn't matter whether $(2,1)$ maps to 11 or 14. – Parcly Taxel Nov 08 '16 at 03:03