Brian Bi

## Section 2.8. Cosets

Exercise 2.8.2 Suppose that $$AX = B$$ has some solution $$v$$. Then we claim that the set of solutions coincides with the coset $$v + W$$. First, for all $$w \in W$$, we have $$A(v+w) = Av + Aw = B + 0 = B$$, so the solution set contains the coset $$v + W$$. For the other direction, if $$v'$$ is a solution, then $$A(v'-v) = Av' - Av = B - B = 0$$, so $$v' - v \in W$$, therefore the solution set is contained in $$v + W$$.

Exercise 2.8.3 Yes. The group contains at least one element other than 1, so it contains some element $$g$$ whose order is not 1. The order of $$g$$ must then be $$p^k$$ for some $$k \geq 1$$ since the order of $$g$$ must divide the order of the group. In that case, the element $$g^{p^{k-1}}$$ has order $$p$$.

Exercise 2.8.4 Yes. Each element of the group other than the identity must have order 5, 7, or 35. If there is an element of order 35, then the group is cyclic and $$x^5$$ and $$x^7$$ have orders 7 and 5, respectively, where $$x$$ is any generator. If the group is not cyclic, then all non-identity elements have order 5 or 7. Now, suppose $$g_1, g_2$$ are two elements of the same order $$p$$, where $$p$$ is prime. Then $$\langle g_1 \rangle \cap \langle g_2 \rangle$$ is a subgroup of $$\langle g_1\rangle$$, so its order is either 1 or $$p$$; that is, $$\langle g_1 \rangle$$ and $$\langle g_2 \rangle$$ are either identical or else have trivial intersection. If a group of order 35 contains only elements of orders 1 and 7, consisting of the union of $$n$$ different cyclic subgroups of order 7, then $$35 = 1 + 6n$$ since each cyclic subgroup contains 6 unique elements as well as the shared identity; there is no such $$n$$, so a contradiction has been reached. Likewise, the impossibility of solving $$35 = 1 + 4n$$ in integers implies the impossibility of a group of order 35 having only elements of orders 1 and 5. We conclude that all groups of order 35 contain both an element of order 5 and an element of order 7.

Exercise 2.8.5 The LCM of 6 and 10, namely 30, must divide $$|G|$$. This is all we can say about $$|G|$$; it could be any multiple of 30, as the cyclic group $$C_{30k}$$ always contains elements of orders 6 and 10, namely $$x^{5k}$$ and $$x^{3k}$$ where $$x$$ is any generator.

Exercise 2.8.6 Let $$n$$ be the order of the kernel. Then $$n \divides |G|$$ and $$|\im \varphi| = 18/n$$; furthermore, $$|\im \varphi| \divides |G'|$$. Since we are told $$\varphi$$ is not trivial, $$\im \varphi$$ must have order 3, 5, or 15. But of these, only 3 divides 18, implying $$n = 6$$.

Exercise 2.8.7 (Thanks to Tom Guo for providing this solution.) The element $$x$$ must have order 2, 11, or 22. It can't have order 22 because then $$y$$ would be a power of $$x$$. Let $$H$$ be the subgroup generated by $$x$$ and $$I$$ the subgroup generated by $$x$$ and $$y$$. If $$x$$ has order 2, and $$y$$ is not a power of $$x$$, then $$|I| > 2$$. Since $$|I| \divides 22$$, it must be that $$|I| = 11$$ or $$|I| = 22$$. But $$H$$ is a subgroup of $$I$$, so $$2 \divides |I|$$, therefore $$|I| = 22$$ and $$I = G$$. In the case where $$x$$ has order 11, again, since $$y \notin H$$, we have that the order of $$I$$ is strictly greater than 11, so it can only be 22.

Exercise 2.8.8 All non-identity elements of $$G$$ have order either 5 or 25. If $$G$$ has an element $$x$$ of order 25, then $$x^5$$ has order 5, therefore $$G$$ always contains some element of order 5, and therefore a (cyclic) subgroup of order 5. If $$G$$ has only one subgroup of order 5, then all elements not in that subgroup must have order 25 otherwise they would generate their own subgroup of order 5. So in that case $$G$$ would be cyclic.

Exercise 2.8.9 If $$x, y \in G$$, then $$\varphi(xy) = (xy)^2 = xyxy$$ and $$\varphi(x) \varphi(y) = xxyy$$, and these two are equal iff $$xy = yx$$. Therefore $$\varphi$$ will be an endomorphism iff $$G$$ is abelian. By a simple pairing argument (see Problem 2.M.2(a)), we see that every group of even order contains an element of order 2, while the converse follows from Lagrange's theorem. Therefore, if $$G$$ has even order then $$\varphi$$ can't be an automorphism (it will map some non-identity element to the identity) whereas if $$G$$ has odd order then no non-identity element will be mapped to the identity, making $$\varphi$$ an automorphism. In conclusion, the finite groups for which $$\varphi$$ is an automorphism are precisely the finite abelian groups of odd order.

Exercise 2.8.10 Let $$H$$ be a subgroup of $$G$$ of index 2. Then $$H$$ has two left cosets in $$H$$, one of which is $$H$$ itself, implying the other is $$G\setminus H$$. The same must hold for the right cosets. Therefore the left and right cosets coincide, and $$H$$ is normal. But the subgroup $$\{1, y\}$$ of index 3 in $$S_3$$ is not a normal subgroup, since $$xyx^{-1} = x^2 y \notin \{1, y\}$$.

Exercise 2.8.12 If $$g \in S$$, then The coset $$g^{-1}S$$ contains $$g^{-1}$$ since $$1 \in S$$, and it also contains 1 since $$g \in S$$. Meanwhile, the coset $$1S = S$$ contains both 1 and $$g$$. Since they share a common element, the two cosets coincide, therefore $$1S = S$$ contains $$g^{-1}$$; $$S$$ is closed under inverses. Now let $$a, b \in S$$. Then the coset $$aS$$ contains both $$a$$ since $$1 \in S$$ and also $$ab$$ since $$b \in S$$, therefore $$aS$$ coincides with $$1S = S$$ which contains $$a$$; therefore the latter contains $$ab$$, so $$S$$ is closed under the operation of $$G$$ as well.