Hamermesh: Chapter 1
I recently finished Chapter 1 of Group Theory and Its Application to Physical Problems by Morton Hamermesh. I picked up this book to complement Quantum Computation and Quantum Information by Nielsen and Chuang, which I’m working through in parallel. QCQI introduces several group theory ideas but doesn’t explore them in depth, so I wanted a resource that develops the concepts more systematically. Hamermesh does exactly that; his exposition is slower, richer in examples, and builds intuition before asking the reader to tackle problems.
One small note: the book uses the term invariant subgroup, which is largely outdated; modern texts refer to these as normal subgroups. Otherwise, the treatment has held up surprisingly well. I’ve included the solutions to the problems in chapter 1 below.
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Page 4
Problem 1
Projective transformation of a line is defined by
\[\begin{aligned} x\rightarrow x'; & x'=\frac{ax+b}{cx+d}, & \text{where $ad-bc\neq 0$} \end{aligned}\]and so
\[\begin{aligned} \frac{(x_1-x_2)/(x_3-x_2)}{(x_1-x_4)/(x_3-x_4)} &\rightarrow \frac{\left(\frac{ax_1+b}{cx_1+d}-\frac{ax_2+b}{cx_2+d}\right)/\left(\frac{ax_3+b}{cx_3+d}-\frac{ax_2+b}{cx_2+d}\right)}{\left(\frac{ax_1+b}{cx_1+d}-\frac{ax_4+b}{cx_4+d}\right)/\left(\frac{ax_3+b}{cx_3+d}-\frac{ax_4+b}{cx_4+d}\right)} \\ &= \frac{\left(\frac{ax_1+b}{cx_1+d}-\frac{ax_2+b}{cx_2+d}\right)\left(\frac{ax_3+b}{cx_3+d}-\frac{ax_4+b}{cx_4+d}\right)}{\left(\frac{ax_3+b}{cx_3+d}-\frac{ax_2+b}{cx_2+d}\right)\left(\frac{ax_1+b}{cx_1+d}-\frac{ax_4+b}{cx_4+d}\right)} \\ &= \frac{\left(\frac{(ax_1+b)(cx_2+d)-(ax_2+b)(cx_1+d)}{(cx_1+d)(cx_2+d)}\right)\left(\frac{(ax_3+b)(cx_4+d)-(ax_4+b)(cx_3+d)}{(cx_3+d)(cx_4+d)}\right)}{\left(\frac{(ax_3+b)(cx_2+d)-(ax_2+b)(cx_3+d)}{(cx_3+d)(cx_2+d)}\right)\left(\frac{(ax_1+b)(cx_4+d)-(ax_4+b)(cx_1+d)}{(cx_1+d)(cx_4+d)}\right)}\\ &= \frac{\left((ax_1+b)(cx_2+d)-(ax_2+b)(cx_1+d)\right)\left((ax_3+b)(cx_4+d)-(ax_4+b)(cx_3+d)\right)}{\left((ax_3+b)(cx_2+d)-(ax_2+b)(cx_3+d)\right)\left((ax_1+b)(cx_4+d)-(ax_4+b)(cx_1+d)\right)}\\ &= \frac{\left((ad-bc)(x_1-x_2)\right)\left((ad-bc)(x_3-x_4)\right)}{\left((ad-bc)(x_3-x_2)\right)\left((ad-bc)(x_1-x_4)\right)}\\ &= \frac{(x_1-x_2)(x_3-x_4)}{(x_3-x_2)(x_1-x_4)}\\ &= \frac{(x_1-x_2)/(x_3-x_2)}{(x_1-x_4)/(x_3-x_4)} \end{aligned}\]Therefore, the cross ratio is invariant under projective transformation.
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Page 11
Problem 1
For the two groups given $a^2 = b$ and $a^2 = b^2 = c^2 = e$. We know $a^2 \neq a$ because then $a=e$. If we said $a^2 = c$ then we would get a group of the same form as (A), just with $b$ and $c$ swapped so this would not be a distinct structure. There are no other elements of the group that we could say $a^2$ is equal to, therefore there are no other distinct structures for a group of order 4.
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Problem 2
Group (A) is $\lbrace e, a, a^2, a^3 \rbrace$ and group (B) is $\lbrace e, a, b, c \rbrace$. To show that a group is abelian, we need to show that $ab = ba$ for all $a,b$ in the group.
For group (A), since it is cyclic it is abelian.
For group (B), for any $a,b$ in the group, since $a^2=e$ and $b^2=e$, then $a=a^{-1}$ and $b=b^{-1}$. Therefore, $(ab)^{-1} = b^{-1}a^{-1} = ba$. But since $(ab)^2 = c^2 = e$, then also $(ab)^{-1} = ab$ and so $ab=(ab)^{-1} = ba$. Therefore, group (B) is abelian.
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Problem 3
For group (A), a realization could be $0^\circ$, $90^\circ$, $180^\circ$, and $270^\circ$ rotation about a fixed axis in three dimensions.
For group (B), a realization could be no rotation, $180^\circ$ rotation about the x-axis, $180^\circ$ rotation about the y-axis, and $180^\circ$ rotation about the z-axis in three dimensions.
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Page 20
Problem 1
The elements are $e$, $(123456)$, and the powers of $(123456)$ which are $(135)(246)$, $(14)(25)(36)$, $(153)(264)$, and $(165432)$.
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Problem 2
This problem is equivalent to finding the structure of the regular subgroups of $S_6$. For regular permutation subgroups, we know that all the cycles in a permutation must have the same length. Therefore, we know that the elements in a subgroup must have either one 6-cycle, two 3-cycles, or three 2-cycles.
The elements of a subgroup containing a 6-cycle are given in problem (1) which is a cyclic group of order 6, $C_6$.
The order of two 3-cycles can be found by taking powers of an example permutation, say $(123)(456)$, which are $(132)(465)$, and $e$. Therefore, two 3-cycles are of order 3.
The order of three 2-cycles can also be found by taking powers of an example permutation, say $(15)(24)(36)$, when you square this permutation, you get $e$, so three 2-cycles are of order 2.
Therefore, every non-identity element must be one of:
- a single 6-cycle (order 6)
- a product of two disjoint 3-cycles (order 3)
- a product of three disjoint 2-cycles (order 2)
Looking at the different permutations of the two 3-cycle and three 2-cycles elements, we get
\[\begin{aligned} \lbrace e, (123)(456), (132)(465), (15)(24)(36), (14)(26)(35), (16)(25)(34) \rbrace \end{aligned}\]which is of order 6. If we set $(123)(456)=a$ and $(15)(24)(36)=b$, we see that this group is equivalent to $\lbrace e, a, a^2, b, ba, ba^2 \rbrace$ with $bab=a^{-1}$ and so this group is isomorphic to $S_3$
Therefore, for groups that are order 6, if the group has an element of order 6 it is $C_6$; otherwise, it is $S_3$.
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Page 22
Problem 1
By taking successive powers of $(1234)$ we get the cyclic subgroup of $S_4$, $H=\lbrace (1234), (13)(24), (1432), e \rbrace$. Let’s calculate some left cosets.
\[\begin{aligned} eH &= \lbrace (1234), (13)(24), (1432), e \rbrace = H \\ (12)H &= \lbrace (234), (1324), (143), (12) \rbrace \\ (13)H &= \lbrace (12)(34), (24), (14)(23), (13) \rbrace \\ (14)H &= \lbrace (123), (1342), (243), (14) \rbrace \\ (23)H &= \lbrace (134), (1243), (142), (23) \rbrace \\ (24)H &= \lbrace (14)(23), (13), (12)(34), (24) \rbrace = (13)H \\ (34)H &= \lbrace (124), (1423), (132), (34) \rbrace \\ \end{aligned}\]The number of elements in $S_4 = 4! = 24$. Therefore, we can define
\[\begin{aligned} S_4 = H + (12)H + (13)H + (14)H + (23)H + (34)H, \end{aligned}\]which has 24 unique elements.
Now let’s do the same for right cosets.
\[\begin{aligned} He &= \lbrace (1234), (13)(24), (1432), e \rbrace = H \\ H(12) &= \lbrace (134), (1423), (243), (12) \rbrace \\ H(13) &= \lbrace (14)(23), (24), (12)(34), (13) \rbrace \\ H(14) &= \lbrace (234), (1243), (132), (14) \rbrace \\ H(23) &= \lbrace (124), (1342), (143), (23) \rbrace \\ H(24) &= \lbrace (12)(34), (13), (14)(23), (24) \rbrace = H(13)\\ H(34) &= \lbrace (123), (1324), (142), (34) \rbrace \\ \end{aligned}\]Therefore, we can define
\[\begin{aligned} S_4 = H + H(12) + H(13) + H(14) + H(23) + H(34). \end{aligned}\]This is a similar resolution as in the same permutations are multiplied to $H$, however the left and right cosets are not equal.
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Page 23
Problem 1
If the group contains an element $a$ of order 8, then the group is a cyclic group $\lbrace a, a^2, a^3, a^4, a^5, a^6, a^7, a^8=e \rbrace$.
To find other possible structure, suppose that the group contains no elements of order 8, but instead has element $a$ of order 4. Thus, the group contains the subgroup $\lbrace a, a^2, a^3, a^4=e \rbrace$. If the group contains another distinct element, b, then it contains 8 distinct elements $\lbrace e, a, a^2, a^3, b, ba, ba^2, ba^3 \rbrace$.
The element $b$ has order 2 or 4. If $b$ is order 4, then the element $b^3$ and $b^2$ must each be equal to one of the 8 distinct elements. Let’s consider the options:
- $b^2=e$ or $b^3=e$: neither of these work because we assumed that $b$ has order 4
- $b^3=a$ or $b^2=a$: if $b^3=a$ that implies $ba=e$ and if $b^2=a$ that impiles $a^2=e$, these contradict the assumtion that $ba$ and $a^2$ are distinct elements.
- $b^3=a^2$: if $b^3=a^2$ it implies $ba^2=e$ which contradicts the assumption that $ba$ is a distinct element.
- $b^3=a^3$ or $b^2=a^3$: if $b^3=a^3$ it implies $ba^3=e$ and if $b^2=a^3$ it implies $e = a^2$ these contradict the assumption that $ba^3$ and $a^2$ are distinct elements.
- $b^3=b$ or $b^2=b$: if $b^3=b$ that implies $b^2=e$ which we already domonstrated is not possible and if $b^2=b$ it implies that $b=e$ which contradicts the assumption that $b$ is a distinct element.
- $b^3=ba$ or $b^2=ba$: if $b^3=ba$ it implies that $a=b^2$ which we already demonstrated is not possible and if $b^2=ba$ it implies that $b=a$ which contradicts the assumption that $b$ is a distinct element.
- $b^2=ba^2$: if $b^2=ba^2$ it implies that $b=a^2$ which contradicts the assumption that $b$ is a distinct element.
- $b^3=ba^3$ or $b^2=ba^3$: if $b^3=ba^3$ it implies that $b^2=a^3$ which we already demonstrated is not possible and if $b^2=ba^3$ that implies $b=a^3$ which contradicts the assumption that $b$ is a distinct element.
Looking at what is left we see
- $b^2=a^2$
- $b^3=ba^2$
Therefore, $b$ can have order 4 if and only if $b^2=a^2$. Otherwise, $b$ has order 2 and $b^2=e$.
Now let’s check to see if the elements $a$ and $b$ can commute. In the book, the authors check the order of $ab$ when determining the possible structure of groups of order 6 to see if it contradicts any assumptions made thus far and so we’ll perform a similar check for groups of order 8. If $a$ and $b$ commute $ab = ba$ and so $(ab)^2 = (ab)(ab) = (ab)(ba) = ab^2a$. When $b$ is order 4, $ab^2a = a^4=e$, which does not contradict any assumptions. When $b$ is order 2, $ab^2a = a^2 \neq e$. Checking larger powers, we see $(ab)^3 = ba^3$, $(ab)^4 = e$ and so $(ab)$ is order 4, which also does not contradict any assumptions. So unlike groups of order 6, non-cyclic groups of order 8 can have the elements commute.
Let’s now suppose the group contains no order 4 or order 8 elements. Now we have $a$, $b$, and $c$ all of order 2. Then the elements are $\lbrace e, a, b, c, ab, ac, bc, abc \rbrace$. Since all the elements are order 2, $(ab)^2=(bc)^2=(ca)^2=e$ and so, using $ab$ as an example, $(ab)^2=abab = e \Rightarrow ab=ba$. Therefore, all terms must commute.
Therefore, the following groups are possible for groups of order 8:
- $\braket{a \vert a^8=e}$
- $\braket{a,b \vert a^4=b^4=e, a^2=b^2, bab^{-1} = a}$
- $\braket{a,b \vert a^4=b^2=e, bab^{-1} = a}$
- $\braket{a,b \vert a^4=b^4=e, a^2=b^2, bab^{-1} = a^{-1}}$
- $\braket{a,b \vert a^4=b^2=e, bab^{-1} = a^{-1}}$
- $\braket{a,b,c \vert a^2=b^2=c^2=e, \text{all commute}}$
The group table for $\braket{a \vert a^8=e}$ is
| $e$ | $a$ | $a^2$ | $a^3$ | $a^4$ | $a^5$ | $a^6$ | $a^7$ |
|---|---|---|---|---|---|---|---|
| $a$ | $a^2$ | $a^3$ | $a^4$ | $a^5$ | $a^6$ | $a^7$ | $e$ |
| $a^2$ | $a^3$ | $a^4$ | $a^5$ | $a^6$ | $a^7$ | $e$ | $a$ |
| $a^3$ | $a^4$ | $a^5$ | $a^6$ | $a^7$ | $e$ | $a$ | $a^2$ |
| $a^4$ | $a^5$ | $a^6$ | $a^7$ | $e$ | $a$ | $a^2$ | $a^3$ |
| $a^5$ | $a^6$ | $a^7$ | $e$ | $a$ | $a^2$ | $a^3$ | $a^4$ |
| $a^6$ | $a^7$ | $e$ | $a$ | $a^2$ | $a^3$ | $a^4$ | $a^5$ |
| $a^7$ | $e$ | $a$ | $a^2$ | $a^3$ | $a^4$ | $a^5$ | $a^6$ |
The group table for $\braket{a,b \vert a^4=b^4=e, a^2=b^2, bab^{-1} = a}$ is
| $e$ | $a$ | $a^2$ | $a^3$ | $b$ | $ba$ | $ba^2$ | $ba^3$ |
|---|---|---|---|---|---|---|---|
| $a$ | $a^2$ | $a^3$ | $e$ | $ba$ | $ba^2$ | $ba^3$ | $b$ |
| $a^2$ | $a^3$ | $e$ | $a$ | $ba^2$ | $ba^3$ | $b$ | $ba$ |
| $a^3$ | $e$ | $a$ | $a^2$ | $ba^3$ | $b$ | $ba$ | $ba^2$ |
| $b$ | $ba$ | $ba^2$ | $ba^3$ | $a^2$ | $a^3$ | $e$ | $a$ |
| $ba$ | $ba^2$ | $ba^3$ | $b$ | $a^3$ | $e$ | $a$ | $a^2$ |
| $ba^2$ | $ba^3$ | $b$ | $ba$ | $e$ | $a$ | $a^2$ | $a^3$ |
| $ba^3$ | $b$ | $ba$ | $ba^2$ | $a$ | $a^2$ | $a^3$ | $e$ |
The group table for $\braket{a,b \vert a^4=b^2=e, bab^{-1} = a}$ is
| $e$ | $a$ | $a^2$ | $a^3$ | $b$ | $ba$ | $ba^2$ | $ba^3$ |
|---|---|---|---|---|---|---|---|
| $a$ | $a^2$ | $a^3$ | $e$ | $ba$ | $ba^2$ | $ba^3$ | $b$ |
| $a^2$ | $a^3$ | $e$ | $a$ | $ba^2$ | $ba^3$ | $b$ | $ba$ |
| $a^3$ | $e$ | $a$ | $a^2$ | $ba^3$ | $b$ | $ba$ | $ba^2$ |
| $b$ | $ba$ | $ba^2$ | $ba^3$ | $e$ | $a$ | $a^2$ | $a^3$ |
| $ba$ | $ba^2$ | $ba^3$ | $b$ | $a$ | $a^2$ | $a^3$ | $e$ |
| $ba^2$ | $ba^3$ | $b$ | $ba$ | $a^2$ | $a^3$ | $e$ | $a$ |
| $ba^3$ | $b$ | $ba$ | $ba^2$ | $a^3$ | $e$ | $a$ | $a^2$ |
The group table for $\braket{a,b \vert a^4=b^4=e, a^2=b^2, bab^{-1} = a^{-1}}$ is
| $e$ | $a$ | $a^2$ | $a^3$ | $b$ | $ba$ | $ba^2$ | $ba^3$ |
|---|---|---|---|---|---|---|---|
| $a$ | $a^2$ | $a^3$ | $e$ | $ba^3$ | $b$ | $ba$ | $ba^2$ |
| $a^2$ | $a^3$ | $e$ | $a$ | $ba^2$ | $ba^3$ | $b$ | $ba$ |
| $a^3$ | $e$ | $a$ | $a^2$ | $ba$ | $ba^2$ | $ba^3$ | $b$ |
| $b$ | $ba$ | $ba^2$ | $ba^3$ | $a^2$ | $a^3$ | $e$ | $a$ |
| $ba$ | $ba^2$ | $ba^3$ | $b$ | $a$ | $a^2$ | $a^3$ | $e$ |
| $ba^2$ | $ba^3$ | $b$ | $ba$ | $e$ | $a$ | $a^2$ | $a^3$ |
| $ba^3$ | $b$ | $ba$ | $ba^2$ | $a^3$ | $e$ | $a$ | $a^2$ |
The group table for $\braket{a,b \vert a^4=b^2=e, bab^{-1} = a^{-1}}$ is
| $e$ | $a$ | $a^2$ | $a^3$ | $b$ | $ba$ | $ba^2$ | $ba^3$ |
|---|---|---|---|---|---|---|---|
| $a$ | $a^2$ | $a^3$ | $e$ | $ba^3$ | $b$ | $ba$ | $ba^2$ |
| $a^2$ | $a^3$ | $e$ | $a$ | $ba^2$ | $ba^3$ | $b$ | $ba$ |
| $a^3$ | $e$ | $a$ | $a^2$ | $ba$ | $ba^2$ | $ba^3$ | $b$ |
| $b$ | $ba$ | $ba^2$ | $ba^3$ | $e$ | $a$ | $a^2$ | $a^3$ |
| $ba$ | $ba^2$ | $ba^3$ | $b$ | $a^3$ | $e$ | $a$ | $a^2$ |
| $ba^2$ | $ba^3$ | $b$ | $ba$ | $a^2$ | $a^3$ | $e$ | $a$ |
| $ba^3$ | $b$ | $ba$ | $ba^2$ | $a$ | $a^2$ | $a^3$ | $e$ |
The group table for $\braket{a,b,c \vert a^2=b^2=c^2=e, \text{all commute}}$ is
| $e$ | $a$ | $b$ | $c$ | $ab$ | $bc$ | $ca$ | $abc$ |
|---|---|---|---|---|---|---|---|
| $a$ | $e$ | $ab$ | $ca$ | $b$ | $abc$ | $c$ | $bc$ |
| $b$ | $ab$ | $e$ | $bc$ | $a$ | $c$ | $abc$ | $ca$ |
| $c$ | $ca$ | $bc$ | $e$ | $abc$ | $b$ | $a$ | $ab$ |
| $ab$ | $b$ | $a$ | $abc$ | $e$ | $ca$ | $bc$ | $c$ |
| $bc$ | $abc$ | $c$ | $b$ | $ca$ | $e$ | $ab$ | $a$ |
| $ca$ | $c$ | $abc$ | $a$ | $bc$ | $ab$ | $e$ | $b$ |
| $abc$ | $bc$ | $ca$ | $ab$ | $c$ | $a$ | $b$ | $e$ |
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Page 25
Problem 1
In $S_5$ the distinct classes are:
\[\begin{aligned} & e \\ & (12),(13),(14),(15),(23),(24),(25),(34),(35),(45) \\ & (123),(124),(125),(132),(142),(152),(134),(135),(143),(153),(145),(154),(234),(235),(243),(253),(245),(254),(345),(354) \\ & (123)(45),(132)(45),(124)(35),(142)(35),(125)(34),(152)(34),(134)(25),(143)(25),(135)(24),(153)(24),(145)(23),(154)(23),(234)(15),(243)(15),(235)(14),(253)(14),(245)(13),(254)(13),(345)(12),(354)(12) \\ & (1234),(1235),(1243),(1253),(1245),(1254),(1324),(1325),(1342),(1352),(1423),(1523),(1432),(1532),(1425),(1452),(1524),(1542),(2345),(2354),(2435),(2453),(2534),(2543),(1345),(1354),(1435),(1453),(1534),(1543) \\ & (12345),(12354),(12435),(12453),(12534),(12543),(13245),(13254),(13425),(13452),(13524),(13542),(14235),(14253),(14325),(14352),(14523),(14532),(15234),(15243),(15324),(15342),(15423),(15432) \\ \end{aligned}\]| Back to top | Solutions Index | Blog Archive |
Page 27
Problem 1
The table through $n=7$ is
\[\begin{aligned} S_1: & (1); & r=1, \\ S_2: & (2), (1^2); & r=2 \\ S_3: & (3), (21), (1^3); & r=3 \\ S_4: & (4), (31), (2^2), (21^2), (1^4); & r=5 \\ S_5: & (5), (41), (32), (31^2), (2^21), (21^3), (1^5); & r= 7 \\ S_6: & (6), (51), (42), (41^2), (3^2), (321), (31^3), (2^3), (2^21^2), (21^4), (1^6); & r= 11\\ S_7: & (7), (61), (52), (51^2), (43), (421), (41^3), (3^21), (32^2), (321^2), (31^4), (2^31), (2^21^3), (21^5), (1^7); & r= 15 \end{aligned}\]In the book, these partitions for $S_5$ are of the form $(\lambda_1 \lambda_2 \lambda_3 \lambda_4 \lambda_5)$, where the number of $j-\text{cycles}$ is given by $v_j = \lambda_j-\lambda_{j+1}$ for $j<5$ and $v_5=\lambda_5$. Therefore, the structure for each partition for $S_5$ is
\[\begin{aligned} \text{partition (book notation)} & (\lambda_1 \lambda_2 \lambda_3 \lambda_4 \lambda_5) & (v_1 v_2 v_3 v_4 v_5) & \text{example} & \text{number of permutations} \\ (5) & (50000) & (50000) & e & 1\\ (41) & (41000) & (31000) & (12) & 10 \\ (32) & (32000) & (12000) & (12)(34) & 15\\ (31^2) & (31100) & (20100) & (123) & 20 \\ (2^21) & (22100) & (01100) & (123)(45) & 20\\ (21^3) & (21110) & (10010) & (1234) & 30\\ (1^5) & (11111) & (00001) & (12345) & 24\\ \end{aligned}\]| Back to top | Solutions Index | Blog Archive |
Page 28
Problem 1
\[\begin{aligned} (1234)(5678) & \text{and} & (1537)(2846) \end{aligned}\]show that one generates a group of order 8. Separate the elements of the group into conjugate classes. Show that this group (the quaternion group) is isomorphic to the group with elements
\[\begin{aligned} 1, -1, i, -i, j, -j, k, -k;\\ \end{aligned}\] \[\begin{aligned} i^2=j^2=k^2=-1, & ij=k, & jk=i, & ki=j \end{aligned}\]Let $a=(1234)(5678)$ and $b=(1537)(2846)$ then
\[\begin{aligned} a^2 &= (13)(24)(57)(68) & &\Leftrightarrow i^2 = -1\\ b^2 &= (13)(24)(57)(68) &= a^2 &\Leftrightarrow j^2 = -1\\ ab &= (1638)(2547) &= c &\Leftrightarrow ij=k\\ c^2 &= (13)(24)(57)(68) &= a^2 = b^2 &\Leftrightarrow k^2 = -1\\ bc &= (1234)(5678) &= a &\Leftrightarrow jk = i\\ ca &= (1537)(2846) &= b &\Leftrightarrow ki = j\\ a^4 &= e & = b^4 = c^4 &\Leftrightarrow 1 = i^4=j^4=k^4\\ \end{aligned}\]Therefore, the group $\lbrace e, a^2, a, a^3, b, b^3, ab, (ab)^3\rbrace$ is isomorphic to $\lbrace 1, -1, i, -i, j, -j, k, -k \rbrace$.
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Page 29
Problem 1
The subgroups are invariant if they contain either all or none of the elements in a given class of $S_4$. None of these subgroups are invariant.
The conjugate subgroups are calculated from $aHa^{-1}$ where $a$ is any element in $S_4$ and $H$ are the subgroups of $S_4$. When we form $aHa^{-1}$ with elements $a$ from $S_4$, we must get cycle structures that are the same as those of elements of $H$ and a subgroup that is the same size as $H$.
For $S_3$ each conjugate subgroup will be of the structure $\lbrace e, (12), (13), (23), (123), (132) \rbrace$, with just permutations of the numbers $\lbrace 1,2,3,4 \rbrace$. There will be four such conjugates.
For $S_2$ each conjugate subgroup with be of the structure $\lbrace e, (12) \rbrace$, with just permutations of the numbers $\lbrace 1,2,3,4 \rbrace$. There will be six such conjugates.
Similarly, for the cyclic group each conjugate subgroup with be of the structure $\lbrack e, (123), (132) \rbrack$, with just permutations of the numbers $\lbrace 1,2,3,4 \rbrace$. There will be four such conjugates.
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Problem 2
The index of a subgroup is the number of distinct cosets of $H$ there are in $G$. If the index is 2, there are only two distinct left cosets in $G$: $H$ and $aH$ for $a\notin H$. There are also only two distinct right cosets: $H$ and $Ha$ for $a\notin H$. We see that $aH$ is all elements of $G$ that are not in $H$ and $Ha$ is also all elements of $G$ that are not in $H$. Therefore, $aH=Ha$. So, for any $g\in G$ it is either in $H$ and so $gH=H=Hg$ or it is not in $H$ and so $gH=aH=Ha=Hg$. Thus $gH=Hg$ for all $g$. Therefore $H$ is invariant.
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Problem 3
A quaternion group is defined by $G=\lbrace 1,-1,i,-i,j,-j,k,-k\rbrace$ where $i^2=j^2=k^2=-1$, $ij=k$, $jk=i$, $ki=j$. Since any group with two of the components in $\lbrace i,j,k \rbrace$ will be able to generate the third component, the only nontrivial subgroups only contain one (or none) of them and is therefore of the structure $\lbrace 1, -1, a, -a \rbrace$ where $a=i,j,\text{ or }k$, $\lbrace 1, -1 \rbrace$, or is $\lbrace 1 \rbrace$.
For subgroups $\lbrace 1, -1, a, -a \rbrace$ where $a=i,j,\text{ or }k$, their index is 2 and so by problem (2) are necessarily invariant. For $H=\lbrace 1, -1 \rbrace$ for every $g\in G$, $gHg^{-1}=H$ since all elements in $H$ commute with all $g\in G$ and so $H$ is invariant. The same can be said for $\lbrace 1 \rbrace$. Therefore, all subgroups in the quaternion group are invariant.
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