MATH 301 Midterm Solutions
1. For groups G and H, their direct product
G × H = {(g, h)| g ? G and h ? H}
forms a group under the operation defined by
(g, h)(g
0
, h0
) = (gg0
, hh0
), for all g, g ? G and h, h0 ? H.
(a) Show that every line N passing through the origin in R
2
is a
normal subgroup of G = (R
2
, +).
(b) For any such line N as in (a), describe G/N.
(c) Show that G/N ?= R.
Solution. (a) As R
2
is a direct product R × R, it is a group under
component-wise addition. Any line passing through the origin in R
2
has an equation of the form y = mx, where m denotes the slope of the
line. As a set, such a line is given by
N = {(x, mx)| x ? R}.
Clearly, N ? G, for if r, s ? N, where r = (x, mx) and s = (y, my),
then we have
rs?1 = (x, mx) + (?y, ?my) = (x ? y, m(x ? y)) ? N.
To show that NEG, observe that for any (x
0
, y0
) ? R
2 and (x, mx) ? N,
we have
(x
0
, y0
)(x, mx)(x
0
, y0
)
?1 = (x
0 + x ? x
0
, y0 + mx ? y
0
) = (x, mx) ? N.
This implies that gNg?1 ? N for all g ? G, or in other words, N E G.
(b) By definition, G/N = {g + N | g ? G}, so for g = (x
0
, y0
) and N as
described in (a), the coset g + N is given by
g + N = {(x
0 + x, y0 + mx)| x ? R}.
That is, g + N is the set of all points on the line y ? y
0 = m(x ? x
0
),
which is parallel to N. In general, any line parallel to N has a equation
1
given by (y ? y0) = m(x ? x0), and the set of all points on this line
equals the coset (x0, y0) + N. From this, we conclude that G/N is the
collection of all lines parallel to N.
(c) Consider the map ? : G ? R defined by
?(x, y) = mx ? y, for all (x, y) ? G.
Then clearly, ? is well-defined, for if (x, y) = (x
0
, y0
), then x = x
0 and
y = y
0
, and so
mx ? y = mx0 ? y
0 =? ?((x, y)) = ?((x
0
, y0
)).
Also, ? is a homomorphism, as
?((x, y) + (x
0
y
0
)) = ?(x + x
0
, y + y
0
)
= m(x + x
0
) ? (y + y
0
)
= (mx ? y) + (mx0 ? y
0
)
= ?((x, y)) + ?((x
0
, y0
)).
The surjectivity of ? follows from the fact that for any r ? R, we have
that ?((0, ?r)) = r.
From the First Isomorphism Theorem, we have
G/Ker ? ?= R.
It remains to show that Ker ? = N. But this follows from the definition
of ?, as
(x, y) ? Ker ? ?? ?((x, y)) = 0
?? mx ? y = 0
?? y = mx
?? (x, y) ? N.
2. Let G be group and H, K E G. Then show that
(a) If H ? K, then (G/H)/(K/H) ?= G/K
(b) If G = HK, then G/(H ? K) ?= G/H × G/K.
Solution. (a) This result is popularly known as the Third Isomorphism
Theorem. First, note that K/H E G/H (which is left as an exercise).
To establish the isomorphism, define a map ? : G/H ? G/K given by
?(gH) = gK, for all gH ? G/H.
2
Then ? is well-defined for the following reason
g1H = g2H =? g1g
?1
2 H = H
=? g1g
?1
2 K = K ( since ?(H) = K)
=? g1K = g2K
=? ?(g1H) = ?(g2H).
As
?(g1H g2H) = ?(g1g2 H) (since H E G)
= g1g2 K
= g1K g2K (since K E G)
= ?(g1)?(g2),
? is a homomorphism. Moreover, ? is surjective, as every coset gK ?
G/K is the image of the coset gH ? G/H under ?. Therefore, by the
First Isomorphism Theorem, we have that
(G/H)/Ker ? ?= G/K.
So it remains to show that Ker ? = K/H, but this follows from the
following argument
gH ? Ker ? ?? ?(gH) = K
?? gK = K
?? g ? K
?? gH ? K/H.
Hence the result follows.
(b) We know from class that if H, K E G, then H ? K E G, from which
we can see that G/(H ? K) is a group. Put N = H ? K, and define a
map ? : G ? G/H × G/K given by
?(g) = (gH, gK), for all g ? G.
Note that ? is well-defined, for if g1 = g2, then
g1g
?1
2 = 1 =? (g1g
?1
2 H, g1g
?1
2 K) = (H, K) (since ?(1) = (H, K))
=? g1g
?1
2 H = H and g1g
?1
2 K = K
=? g1H = g2H and g1K = g2K
=? ?(g2) = ?(g2).
3
Moreover, ? is a homomorphism, as
?(g1g2) = (g1g2H, g1g2K)
= ((g1H)(g2H),(g1K)(g2K)) (since H, K E G)
= (g1H, g1K)(g2H, g2K) (by definition of a direct product)
= ?(g1)?(g2).
To show the surjectivity of ?, take any (gH, g0K) ? G/H ×G/K. Since
G = HK, we have that g = hk and g
0 = h
0k
0
for some h, h0 ? H and
k, k0 ? K. Then then fact that H, K E G, would imply that gH = kH
and g
0K = h
0K, and so (gH, g0K) = (kH, h0K). Hence
?(kh0
) = (kh0H, kh0K)
= (kH, h0K) (since H, K E G)
= (gH, g0K),
which proves that ? is surjective.
Finally, by the First Isomorphism Theorem, we have
G/Ker ? ?= G/H × G/K.
Furthermore,
g ? Ker ? ?? (gH, gK) = (H, K)
?? gH = H and gK = K
?? g ? H and g ? K
?? g ? H ? K,
which establishes that Ker ? = H ? K, and hence the result follows.
3. Consider the set Q8 = {±1, ±i, ±j, ±k} having 8 elements with an
operation · satisfying the following relations
i · i = j · j = k · k = ?1
i · j = k, j · k = i, k · i = j
(?1) · (?1) = +1
(a) Show that (Q8, ·) is a group with +1 as its identity element. (Q8
is called the group of quaternions.)
4
(b) Is (Q8, ·) an abelian group? Explain why or why not.
(c) Show that every subgroup of (Q8, ·) is normal.
Solution. (a)&(b) For simplicity, let us denote +1 by 1, and for any
a, b, x ? Q8, we denote x · x by x
2
, and a · b by ab. Since (?1)(?1) = 1,
we have
o(?1) = 2 (i.e. ? 1 = (?1)?1
) and ? 1 = (1)(?1)?1 = (?1)?1
(1),
or in other words
?1 = (1)(?1) = (?1)(1). (1)
Moreover, i
4 = i
2
i
2 = (?1)(?1) = 1, which implies that o(i) | 4, but
since i
2 = ?1, we infer that o(i) = 4. By an analogous argument we
can conclude that
o(±i) = o(±j) = o(±k) = 4. (2)
Since jk = i, we have j
2k = ji, that is, ji = (?1)k, and a similar
argument can be used to derive all other relations of this type, namely
(?1)k = ji = k(?1)
(?1)i = kj = i(?1)
(?1)j = ik = j(?1) (3)
Next, we establish that (?1)x = ?x for all x ? Q8. For the case
when x = ±1, this follows from (1). So we choose x ? Q8 {±1},
say x = i. Suppose that (?1)i = y., Then it is clear that y 6= i (as
this would imply that ?1 = 1), and from (1) we see that y ± 1, i, so
y ? {?i, ±j ± k}. Suppose that y = j, then we have
(?1)i = j
=? i(?1) = j (from (3))
=? i(?1)j = ?1
=? (?1)ij = ?1 (from (3))
=? (?1)k = ?1,
and so k = 1, which gives a contradiction. Hence, y 6= j, and in a
similar fashion we can conclude that y 6= ?j, ±i, and by elimination
5
we can conclude that (?1)i = ?i, and extending this to j, k (using
similar arguments), we have
(?1)i = ?i = k(?1)
(?1)j = ?j = i(?1)
(?1)k = ?k = k(?1) (4)
Finally, since i
2 = j
2 = k
2 = ?1, we have the relations
i
?1 = ?i, j?1 = ?j, k?1 = ?k
From Equations (1) – (4), we see that Q8 is closed under its operation,
and every element in Q8 has a unique inverse. Hence, the relations
defined in the problem extend to a group operation on Q8. (Note that
the associativity of the operation, which has been implicitly assumed
in some of the arguments above, is a cumbersome but easy exercise.)
From the relations in (3) and (4), it is clear that Q8 is non-abelian.
(c) Since |Q8| = 8, by the Lagranges Theorem, any proper subgroup
of Q8 has to be of order 2 or 4. Furthermore, any subgroup of order 4
has index 2 in Q8, and hence has to be normal. So it suffices to show
that every subgroup of order 2 is normal in Q8.
Since any subgroup of order 2 is cyclic, it has to be generated by an
element in Q8 of order 2. We showed above that ?1 is the only element
of order 2. Since it generates subgroup H = {?1, 1}, H is the only
subgroup of order 2, so it suffices to show that H EQ8. For any g ? Q8
and x ? H, we have gxg?1 = 1, if x =, and when x = ?1, we have
g(?1)g
?1 = (?1)gg?1 = ?1 ? H,
which shows that H E G, and the result follows.
4. We know from class that the dihedral group
D8 = hr, si = {1, r, r2
, r3
, s, sr, sr2
, sr3
}
is the group of symmetries of a square, which is generated by a rotation
r by 2?/4 and a reflection s.
(a) Find all subgroups of D8 order 2.
6
(b) Show that D8 has exactly three subgroups of order 4, one of which
is cyclic, while the remaining two are non-cyclic. (Note that this
gives an example of a non-abelian group of order 4.)
(c) Assuming that isomorphic groups possess the same subgroup structure, establish that Q8 is not isomorphic to D8.
Solution. (a) Every subgroup of D8 of order 2 has to be generated
by an element of order 2. The elements in D8 of order 2 are the reflections s, sr, sr2
, and sr3
, and the rotation r
2 by ?. Hence, D8 has 5
distinct subgroups of order 2, namely the subgroups generated by these
elements.
(b) If a subgroup of order 4 is cyclic, then it has to be generated by an
element g ? D8 of order 4. Since r is the only of order 4, we conclude
that
hri = {1, r, r2
, r3
}
is the only cyclic subgroup of D8 of order 4.
Suppose that H is a non-cyclic subgroup of D8 of order 4. Then by
Lagranges Theorem, every non-trivial element g ? H is of order 2 or 4.
If o(g) = 4, then H is cyclic, which contradicts our assumption. Hence,
every non-trivial element of H is of order 2 ( =? r /? H). In other
words, we have the following observation:
Observation: H has to contain 3 distinct elements of order 2.
Before we find all such order 4 subgroups, first note that since o(s) =
(srk
) = 2, so we have (srk
)(srk
) = 1, which implies that
srk = r
?k
s
?1 = r
n?k
s (*)
Using relation (*) and the observation made earlier, we can see that
{1, r2
, s, sr2
} and {1, r2
, sr, sr3
}
are the only other order 4 subgroups.
Exercise: Show that both the subgroups mentioned above are isomorphic to G = Z2 × Z2. (Note that G is called the Klein-4 group.)
(c) We proved earlier, that Q8 has a unique subgroup of order 2, namely
{?1, 1}. But we now know that D8 has five distinct subgroups of order
2, which shows that D8 cannot be isomorphic to Q8.
7
Exercise: Can we conclude the same, using the structure of the order
4 subgroups?
5. (Bonus) Let H = {z ? C | |z| = 1}. Then show that [10]
R/Z ?= H
Solution. First, realize that
Exercise: H ? C
×.
Then we define a map ? : R ? H given by
? = e
i(2?x)
, for all x ? R.
Note that for any x ? R, we can see that
|?(x)| = |e
i(2?x)
| = | cos(2?x) + isin(2?x)| = 1 =? ?(x) ? H.
We will also need to the establish the following:
Exercise: Show that ? is a well-defined surjective homomorphism.
By the First Isomorphism Theorem, we have that
R/Ker ? ?= H. (*)
To complete the argument, note that
s ? Ker ? ?? ?(s) = 1
?? e
i(2?s) = 1
?? cos(2?s) + isin(2?s) = 1
?? cos(2?s) = 1 and sin(2?s) = 0
?? n ? Z,
which shows that Ker ? = Z. Hence, the result now follows from (*).
8MTH 301 Midterm Solutions
1. For groups G and H, their direct product
G × H = {(g, h)| g ? G and h ? H}
forms a group under the operation defined by
(g, h)(g
0
, h0
) = (gg0
, hh0
), for all g, g ? G and h, h0 ? H.
(a) Show that every line N passing through the origin in R
2
is a
normal subgroup of G = (R
2
, +).
(b) For any such line N as in (a), describe G/N.
(c) Show that G/N ?= R.
Solution. (a) As R
2
is a direct product R × R, it is a group under
component-wise addition. Any line passing through the origin in R
2
has an equation of the form y = mx, where m denotes the slope of the
line. As a set, such a line is given by
N = {(x, mx)| x ? R}.
Clearly, N ? G, for if r, s ? N, where r = (x, mx) and s = (y, my),
then we have
rs?1 = (x, mx) + (?y, ?my) = (x ? y, m(x ? y)) ? N.
To show that NEG, observe that for any (x
0
, y0
) ? R
2 and (x, mx) ? N,
we have
(x
0
, y0
)(x, mx)(x
0
, y0
)
?1 = (x
0 + x ? x
0
, y0 + mx ? y
0
) = (x, mx) ? N.
This implies that gNg?1 ? N for all g ? G, or in other words, N E G.
(b) By definition, G/N = {g + N | g ? G}, so for g = (x
0
, y0
) and N as
described in (a), the coset g + N is given by
g + N = {(x
0 + x, y0 + mx)| x ? R}.
That is, g + N is the set of all points on the line y ? y
0 = m(x ? x
0
),
which is parallel to N. In general, any line parallel to N has a equation
1
given by (y ? y0) = m(x ? x0), and the set of all points on this line
equals the coset (x0, y0) + N. From this, we conclude that G/N is the
collection of all lines parallel to N.
(c) Consider the map ? : G ? R defined by
?(x, y) = mx ? y, for all (x, y) ? G.
Then clearly, ? is well-defined, for if (x, y) = (x
0
, y0
), then x = x
0 and
y = y
0
, and so
mx ? y = mx0 ? y
0 =? ?((x, y)) = ?((x
0
, y0
)).
Also, ? is a homomorphism, as
?((x, y) + (x
0
y
0
)) = ?(x + x
0
, y + y
0
)
= m(x + x
0
) ? (y + y
0
)
= (mx ? y) + (mx0 ? y
0
)
= ?((x, y)) + ?((x
0
, y0
)).
The surjectivity of ? follows from the fact that for any r ? R, we have
that ?((0, ?r)) = r.
From the First Isomorphism Theorem, we have
G/Ker ? ?= R.
It remains to show that Ker ? = N. But this follows from the definition
of ?, as
(x, y) ? Ker ? ?? ?((x, y)) = 0
?? mx ? y = 0
?? y = mx
?? (x, y) ? N.
2. Let G be group and H, K E G. Then show that
(a) If H ? K, then (G/H)/(K/H) ?= G/K
(b) If G = HK, then G/(H ? K) ?= G/H × G/K.
Solution. (a) This result is popularly known as the Third Isomorphism
Theorem. First, note that K/H E G/H (which is left as an exercise).
To establish the isomorphism, define a map ? : G/H ? G/K given by
?(gH) = gK, for all gH ? G/H.
2
Then ? is well-defined for the following reason
g1H = g2H =? g1g
?1
2 H = H
=? g1g
?1
2 K = K ( since ?(H) = K)
=? g1K = g2K
=? ?(g1H) = ?(g2H).
As
?(g1H g2H) = ?(g1g2 H) (since H E G)
= g1g2 K
= g1K g2K (since K E G)
= ?(g1)?(g2),
? is a homomorphism. Moreover, ? is surjective, as every coset gK ?
G/K is the image of the coset gH ? G/H under ?. Therefore, by the
First Isomorphism Theorem, we have that
(G/H)/Ker ? ?= G/K.
So it remains to show that Ker ? = K/H, but this follows from the
following argument
gH ? Ker ? ?? ?(gH) = K
?? gK = K
?? g ? K
?? gH ? K/H.
Hence the result follows.
(b) We know from class that if H, K E G, then H ? K E G, from which
we can see that G/(H ? K) is a group. Put N = H ? K, and define a
map ? : G ? G/H × G/K given by
?(g) = (gH, gK), for all g ? G.
Note that ? is well-defined, for if g1 = g2, then
g1g
?1
2 = 1 =? (g1g
?1
2 H, g1g
?1
2 K) = (H, K) (since ?(1) = (H, K))
=? g1g
?1
2 H = H and g1g
?1
2 K = K
=? g1H = g2H and g1K = g2K
=? ?(g2) = ?(g2).
3
Moreover, ? is a homomorphism, as
?(g1g2) = (g1g2H, g1g2K)
= ((g1H)(g2H),(g1K)(g2K)) (since H, K E G)
= (g1H, g1K)(g2H, g2K) (by definition of a direct product)
= ?(g1)?(g2).
To show the surjectivity of ?, take any (gH, g0K) ? G/H ×G/K. Since
G = HK, we have that g = hk and g
0 = h
0k
0
for some h, h0 ? H and
k, k0 ? K. Then then fact that H, K E G, would imply that gH = kH
and g
0K = h
0K, and so (gH, g0K) = (kH, h0K). Hence
?(kh0
) = (kh0H, kh0K)
= (kH, h0K) (since H, K E G)
= (gH, g0K),
which proves that ? is surjective.
Finally, by the First Isomorphism Theorem, we have
G/Ker ? ?= G/H × G/K.
Furthermore,
g ? Ker ? ?? (gH, gK) = (H, K)
?? gH = H and gK = K
?? g ? H and g ? K
?? g ? H ? K,
which establishes that Ker ? = H ? K, and hence the result follows.
3. Consider the set Q8 = {±1, ±i, ±j, ±k} having 8 elements with an
operation · satisfying the following relations
i · i = j · j = k · k = ?1
i · j = k, j · k = i, k · i = j
(?1) · (?1) = +1
(a) Show that (Q8, ·) is a group with +1 as its identity element. (Q8
is called the group of quaternions.)
4
(b) Is (Q8, ·) an abelian group? Explain why or why not.
(c) Show that every subgroup of (Q8, ·) is normal.
Solution. (a)&(b) For simplicity, let us denote +1 by 1, and for any
a, b, x ? Q8, we denote x · x by x
2
, and a · b by ab. Since (?1)(?1) = 1,
we have
o(?1) = 2 (i.e. ? 1 = (?1)?1
) and ? 1 = (1)(?1)?1 = (?1)?1
(1),
or in other words
?1 = (1)(?1) = (?1)(1). (1)
Moreover, i
4 = i
2
i
2 = (?1)(?1) = 1, which implies that o(i) | 4, but
since i
2 = ?1, we infer that o(i) = 4. By an analogous argument we
can conclude that
o(±i) = o(±j) = o(±k) = 4. (2)
Since jk = i, we have j
2k = ji, that is, ji = (?1)k, and a similar
argument can be used to derive all other relations of this type, namely
(?1)k = ji = k(?1)
(?1)i = kj = i(?1)
(?1)j = ik = j(?1) (3)
Next, we establish that (?1)x = ?x for all x ? Q8. For the case
when x = ±1, this follows from (1). So we choose x ? Q8 {±1},
say x = i. Suppose that (?1)i = y., Then it is clear that y 6= i (as
this would imply that ?1 = 1), and from (1) we see that y ± 1, i, so
y ? {?i, ±j ± k}. Suppose that y = j, then we have
(?1)i = j
=? i(?1) = j (from (3))
=? i(?1)j = ?1
=? (?1)ij = ?1 (from (3))
=? (?1)k = ?1,
and so k = 1, which gives a contradiction. Hence, y 6= j, and in a
similar fashion we can conclude that y 6= ?j, ±i, and by elimination
5
we can conclude that (?1)i = ?i, and extending this to j, k (using
similar arguments), we have
(?1)i = ?i = k(?1)
(?1)j = ?j = i(?1)
(?1)k = ?k = k(?1) (4)
Finally, since i
2 = j
2 = k
2 = ?1, we have the relations
i
?1 = ?i, j?1 = ?j, k?1 = ?k
From Equations (1) – (4), we see that Q8 is closed under its operation,
and every element in Q8 has a unique inverse. Hence, the relations
defined in the problem extend to a group operation on Q8. (Note that
the associativity of the operation, which has been implicitly assumed
in some of the arguments above, is a cumbersome but easy exercise.)
From the relations in (3) and (4), it is clear that Q8 is non-abelian.
(c) Since |Q8| = 8, by the Lagranges Theorem, any proper subgroup
of Q8 has to be of order 2 or 4. Furthermore, any subgroup of order 4
has index 2 in Q8, and hence has to be normal. So it suffices to show
that every subgroup of order 2 is normal in Q8.
Since any subgroup of order 2 is cyclic, it has to be generated by an
element in Q8 of order 2. We showed above that ?1 is the only element
of order 2. Since it generates subgroup H = {?1, 1}, H is the only
subgroup of order 2, so it suffices to show that H EQ8. For any g ? Q8
and x ? H, we have gxg?1 = 1, if x =, and when x = ?1, we have
g(?1)g
?1 = (?1)gg?1 = ?1 ? H,
which shows that H E G, and the result follows.
4. We know from class that the dihedral group
D8 = hr, si = {1, r, r2
, r3
, s, sr, sr2
, sr3
}
is the group of symmetries of a square, which is generated by a rotation
r by 2?/4 and a reflection s.
(a) Find all subgroups of D8 order 2.
6
(b) Show that D8 has exactly three subgroups of order 4, one of which
is cyclic, while the remaining two are non-cyclic. (Note that this
gives an example of a non-abelian group of order 4.)
(c) Assuming that isomorphic groups possess the same subgroup structure, establish that Q8 is not isomorphic to D8.
Solution. (a) Every subgroup of D8 of order 2 has to be generated
by an element of order 2. The elements in D8 of order 2 are the reflections s, sr, sr2
, and sr3
, and the rotation r
2 by ?. Hence, D8 has 5
distinct subgroups of order 2, namely the subgroups generated by these
elements.
(b) If a subgroup of order 4 is cyclic, then it has to be generated by an
element g ? D8 of order 4. Since r is the only of order 4, we conclude
that
hri = {1, r, r2
, r3
}
is the only cyclic subgroup of D8 of order 4.
Suppose that H is a non-cyclic subgroup of D8 of order 4. Then by
Lagranges Theorem, every non-trivial element g ? H is of order 2 or 4.
If o(g) = 4, then H is cyclic, which contradicts our assumption. Hence,
every non-trivial element of H is of order 2 ( =? r /? H). In other
words, we have the following observation:
Observation: H has to contain 3 distinct elements of order 2.
Before we find all such order 4 subgroups, first note that since o(s) =
(srk
) = 2, so we have (srk
)(srk
) = 1, which implies that
srk = r
?k
s
?1 = r
n?k
s (*)
Using relation (*) and the observation made earlier, we can see that
{1, r2
, s, sr2
} and {1, r2
, sr, sr3
}
are the only other order 4 subgroups.
Exercise: Show that both the subgroups mentioned above are isomorphic to G = Z2 × Z2. (Note that G is called the Klein-4 group.)
(c) We proved earlier, that Q8 has a unique subgroup of order 2, namely
{?1, 1}. But we now know that D8 has five distinct subgroups of order
2, which shows that D8 cannot be isomorphic to Q8.
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Exercise: Can we conclude the same, using the structure of the order
4 subgroups?
5. (Bonus) Let H = {z ? C | |z| = 1}. Then show that [10]
R/Z ?= H
Solution. First, realize that
Exercise: H ? C
×.
Then we define a map ? : R ? H given by
? = e
i(2?x)
, for all x ? R.
Note that for any x ? R, we can see that
|?(x)| = |e
i(2?x)
| = | cos(2?x) + isin(2?x)| = 1 =? ?(x) ? H.
We will also need to the establish the following:
Exercise: Show that ? is a well-defined surjective homomorphism.
By the First Isomorphism Theorem, we have that
R/Ker ? ?= H. (*)
To complete the argument, note that
s ? Ker ? ?? ?(s) = 1
?? e
i(2?s) = 1
?? cos(2?s) + isin(2?s) = 1
?? cos(2?s) = 1 and sin(2?s) = 0
?? n ? Z,
which shows that Ker ? = Z. Hence, the result now follows from (*).
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