A subgroup H of a group G is normal if and only if for every g in G, gHg^-1 = H. In other words, if conjugating any element of H by any element of G gives us an element that remains in H, then H is a normal subgroup of G.
To determine if H&G and K&G are normal, we need to examine if every element of G conjugates any element in H and K, respectively, without taking them outside of H and K.
For (i) H&G: Since H has only 4 elements, it is a cyclic subgroup of G generated by (a, b), where a is an element of Z_6 and b is an element of S_3. We can list all the elements of H as {(0, id), (3, id), (1, (1 2)), (4, (1 2))}, where id is the identity element of S_3.
Now, for any (x, y) in G, we need to check if it conjugates any of these 4 elements out of H. Let's suppose (x, y) does take some element outside of H. Then (x, y)(a, b)(x, y)^-1 = (xa(x^-1), yby^-1) must not lie in H. But since G is the direct product of Z_6 and S_3, this means xa(x^-1) must be in Z_6, and yby^-1 must be in S_3.
Since H is a subgroup of G, xa(x^-1) and yby^-1 must be elements of H as well. Therefore, we can conclude that H is a normal subgroup of G because for any (x, y) in G, conjugating any element of H by (x, y) gives us an element that remains in H.
For (ii) K&G: Since |K| = 9 is not a factor of |G| = 2^2 * 3^2, K cannot be a subgroup of G. Therefore, K&G is not even a well-defined subset of G, and we cannot determine if it is a normal subgroup.
Let group G= Z_6 × S_3 and H, K be 2 subgroups of G such that |H|=4 and |K|=9
I know |G|=2^2 * 3^2
What is the normality between the groups
(i) H & G
(ii) K & G
I know that a group is normally iffy the no of P-SSG of it is unique.
What does unique here means?
How do we find if H&G and K&G are normal?
3 answers
Order is G is 36, how can K not be a subgroup of it when order of K is 9?
I apologize for the confusion in my previous response. You are correct that if the order of a subgroup K of a group G is a factor of the order of G, then K must be a subgroup of G.
In this case, since the order of K is 9 and the order of G is 36, K is indeed a subgroup of G. To determine if K is a normal subgroup of G, we need to check if for every g in G, gKg^-1 = K.
Let's first count the number of elements in K. By Lagrange's theorem, the order of every subgroup divides the order of the group. Since the order of K is 9 and the order of G is 36, the only possibilities for the order of K are 1, 3, and 9. K cannot have order 1 since it contains the identity element, and it cannot have order 3 since 3 does not divide 9. Therefore, K must have order 9 and G/K must have order 4 by Lagrange's theorem.
Since the order of G/K is 4, it must be isomorphic to either Z_4 or Z_2 × Z_2. Let's consider each case separately:
Case 1: G/K is isomorphic to Z_4.
In this case, there exists some k in K such that K = {0K, kK, 2kK, 3kK}. Let's suppose that gK = aK is some other coset in G/K. Then we can write g = ak + x, where x is some element of K. Now, for any h in K, we have
ghg^-1 = (ak + x)h(k^-1a^-1 + x') = akhk^-1a^-1 + xhx'.
Since K is a subgroup of G, hk^-1 is also in K. Therefore, ghg^-1 belongs to the coset aK if and only if xhx' is in K. But since K is a subgroup of index 4 in G, there are only 4 cosets, and each element of G belongs to one of these cosets. Therefore, we can conclude that K is a normal subgroup of G.
Case 2: G/K is isomorphic to Z_2 × Z_2.
In this case, there exists some k in K such that K = {(0,0)K, (0,1)K, (1,0)K, (1,1)K}. Let's suppose that gK = (a,b)K is some other coset in G/K. Then we can write g = (ak, bk) + (x,y), where (x,y) is some element of K. Now, for any h in K, we have
ghg^-1 = (ak, bk)(h,0)(k^-1a^-1, k^-1b^-1)(x,y)(kak^-1, kbk^-1) = (ak, bk)h(k^-1a, k^-1b)(kxk^-1 + akxk^-1 - ax, bxyk^-1)
Since K is a subgroup of G, hk^-1 is also in K. Therefore, ghg^-1 belongs to the coset (a,b)K if and only if kxk^-1 + akxk^-1 - ax is 0 (mod 2) and bxyk^-1 is 0 (mod 2). But since K is a subgroup of index 4 in G, there are only 4 cosets, and each element of G belongs to one of these cosets. Therefore, we can conclude that K is a normal subgroup of G.
Hence, we have shown that K is a normal subgroup of G in both cases, and therefore, K is a normal subgroup of G.
In this case, since the order of K is 9 and the order of G is 36, K is indeed a subgroup of G. To determine if K is a normal subgroup of G, we need to check if for every g in G, gKg^-1 = K.
Let's first count the number of elements in K. By Lagrange's theorem, the order of every subgroup divides the order of the group. Since the order of K is 9 and the order of G is 36, the only possibilities for the order of K are 1, 3, and 9. K cannot have order 1 since it contains the identity element, and it cannot have order 3 since 3 does not divide 9. Therefore, K must have order 9 and G/K must have order 4 by Lagrange's theorem.
Since the order of G/K is 4, it must be isomorphic to either Z_4 or Z_2 × Z_2. Let's consider each case separately:
Case 1: G/K is isomorphic to Z_4.
In this case, there exists some k in K such that K = {0K, kK, 2kK, 3kK}. Let's suppose that gK = aK is some other coset in G/K. Then we can write g = ak + x, where x is some element of K. Now, for any h in K, we have
ghg^-1 = (ak + x)h(k^-1a^-1 + x') = akhk^-1a^-1 + xhx'.
Since K is a subgroup of G, hk^-1 is also in K. Therefore, ghg^-1 belongs to the coset aK if and only if xhx' is in K. But since K is a subgroup of index 4 in G, there are only 4 cosets, and each element of G belongs to one of these cosets. Therefore, we can conclude that K is a normal subgroup of G.
Case 2: G/K is isomorphic to Z_2 × Z_2.
In this case, there exists some k in K such that K = {(0,0)K, (0,1)K, (1,0)K, (1,1)K}. Let's suppose that gK = (a,b)K is some other coset in G/K. Then we can write g = (ak, bk) + (x,y), where (x,y) is some element of K. Now, for any h in K, we have
ghg^-1 = (ak, bk)(h,0)(k^-1a^-1, k^-1b^-1)(x,y)(kak^-1, kbk^-1) = (ak, bk)h(k^-1a, k^-1b)(kxk^-1 + akxk^-1 - ax, bxyk^-1)
Since K is a subgroup of G, hk^-1 is also in K. Therefore, ghg^-1 belongs to the coset (a,b)K if and only if kxk^-1 + akxk^-1 - ax is 0 (mod 2) and bxyk^-1 is 0 (mod 2). But since K is a subgroup of index 4 in G, there are only 4 cosets, and each element of G belongs to one of these cosets. Therefore, we can conclude that K is a normal subgroup of G.
Hence, we have shown that K is a normal subgroup of G in both cases, and therefore, K is a normal subgroup of G.
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