1. Find all subgroups of a cyclic group G = of order 12.
2. Prove that the group U(11) is cyclic. List all of the generators for U(11).
3.Suppose G is a group that has exactly 8 elements of order 3. Prove that G has exactly 4 subgroups of order3.
4. Prove that if a group G contains an element g such that ga = g for all a in G, then the group G has just one element.
5. Suppose that g and h are group elements such that |g| = 2 (|g| = order of the element g), h is not the identity and ghg = h^2. Find |h| and prove your answer is correct.
2. Prove that the group U(11) is cyclic. List all of the generators for U(11).
3.Suppose G is a group that has exactly 8 elements of order 3. Prove that G has exactly 4 subgroups of order3.
4. Prove that if a group G contains an element g such that ga = g for all a in G, then the group G has just one element.
5. Suppose that g and h are group elements such that |g| = 2 (|g| = order of the element g), h is not the identity and ghg = h^2. Find |h| and prove your answer is correct.
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1. This is easy, but a little tedious. All subgroups of a cyclic group are cyclic. You can calculate the cyclic group of any element manually by computing powers of it until you get back to e. For example:
= {g^3, g^6, g^9, e}
= {g^5, g^10, g^3, g^8, g, ...} = G ... (once you get back to a confirmed generator like g, you know everything else is in the cyclic group)
Compute all cyclic subgroups generated by each element, cross out any repeats, and this will be a complete list of all subgroups. Including the full group and trivial group, there should be 6 in total.
2. What is U(10)? I looked it up, and all I could find was the group of 11x11 unitary matrices, which is definitely not cyclic, as it is uncountable.
3. To say an element has order 3 is to say the cyclic group generated by this element is of order 3. So, if g in G is an element of order 3, then {g, g^2, e} is a subgroup of order 3. Notice here that g^2 = g^-1 will also generate this subgroup. Therefore, if g is of order 3, so is g^-1. We can therefore form pairs between the order 3 elements by whether they are each other's inverse. If we have exactly 8 elements of order 3, then that gives us 4 pairs.
Now, for each of these pairs, if we add in the identity, we get a subgroup of order 3. Thus, the number of subgroups generated in this way will be exactly 4. Will there be any others of order 3 (i.e. not made by inverse pairs of order 3 elements and the identity)? No, because order 3 subgroups have prime order, and thus are cyclic by a corollary to Lagrange's theorem. Therefore, they are generated by (at least one) element of order 3, so they all look like {g, g^-1, e}. Thus, there are exactly 4 subgroups of order 3.
Compute all cyclic subgroups generated by each element, cross out any repeats, and this will be a complete list of all subgroups. Including the full group and trivial group, there should be 6 in total.
2. What is U(10)? I looked it up, and all I could find was the group of 11x11 unitary matrices, which is definitely not cyclic, as it is uncountable.
3. To say an element has order 3 is to say the cyclic group generated by this element is of order 3. So, if g in G is an element of order 3, then {g, g^2, e} is a subgroup of order 3. Notice here that g^2 = g^-1 will also generate this subgroup. Therefore, if g is of order 3, so is g^-1. We can therefore form pairs between the order 3 elements by whether they are each other's inverse. If we have exactly 8 elements of order 3, then that gives us 4 pairs.
Now, for each of these pairs, if we add in the identity, we get a subgroup of order 3. Thus, the number of subgroups generated in this way will be exactly 4. Will there be any others of order 3 (i.e. not made by inverse pairs of order 3 elements and the identity)? No, because order 3 subgroups have prime order, and thus are cyclic by a corollary to Lagrange's theorem. Therefore, they are generated by (at least one) element of order 3, so they all look like {g, g^-1, e}. Thus, there are exactly 4 subgroups of order 3.
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