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$\mathbb Z / 15\mathbb Z$ is a ring, so when you examine the (possible) invertible elements, you are talking about the unitgroup or multiplicative group.
Now consider $ U(15) $. We want to find automorphism group of $ U(15) $. Generating set of $ U(15) $ is $ \{2, 11\} $ as well as $ \{7, 4\} $ (just to name a few). For $ φ(2) $ there are $ 4 $ options: 2, 7, 8, 13 and for $ φ(11) $ there are $ 3 $ options: $ 4, 11, 14 $. However, the first element is actually the square of any of the order ...
By the Chinese remainder theorem we have that: U15 ≃U3 ×U5 U 15 ≃ U 3 × U 5. hence any element of U15 U 15 has order 1, 2 1, 2 or 4 4. Since |U15| = φ(15) = 8 | U 15 | = φ (15) = 8, there is no element of U15 U 15 that generates it, hence U15 U 15 is not a cyclic group. Share.
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11. Yes, it makes sense. The order of an element g g in some group is the least positive integer n n such that gn = 1 g n = 1 (the identity of the group), if any such n n exists. If there is no such n n, then the order of g g is defined to be ∞ ∞. As noted in the comment by @Travis, you can take a small permutation group to get an example.
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) ↦ 1. (2, 1) ↦ 11. (1, 4) ↦ 4. (2, 4) ↦ 14. 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. What you have done is a little bit weird. The binary operation in G G is multiplication but you are considering H H as an additive subgroup. Actually, we can see without any calculation that H = G H = G, just observing that gcd(15, 4) = 1 g c d (15, 4) = 1. Share.
0. Hint: This group is isomorphic to G =Z4 ×Z2 G = Z 4 × Z 2, in which the group operation is much easier to understand. Find all nontrivial cyclic subgroups of G G. Then find all two-generated subgroups of G G. Then all three-generated subgroups of G G. Keep going until you stop getting new groups (or, for a slightly more clever approach ...
No. Keep in mind that the order of this group is 4 4, so by Lagrange's Theorem the order of every element must be a factor of 4 4 (either 1 1, 2 2, or 4 4). Remember that the group operation is multiplication modulo 10 10, not addition modulo 10 10. So to find the order of 3 3, compute 3k (mod 10) 3 k (mod 10) for k = 1, 2, 4 k = 1, 2, 4 until ...
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