a.) How many subgroups of order 4 does D4 have? Justify your answer. b.) Let G b
ID: 3113728 • Letter: A
Question
a.) How many subgroups of order 4 does D4 have? Justify your answer.b.) Let G be a finite group with more than one element. Show that G has an element of prime order.
a.) How many subgroups of order 4 does D4 have? Justify your answer.
b.) Let G be a finite group with more than one element. Show that G has an element of prime order.
b.) Let G be a finite group with more than one element. Show that G has an element of prime order. b.) Let G be a finite group with more than one element. Show that G has an element of prime order.
Explanation / Answer
a)order of Dn we all know is 2*n so for D4 order is 2*4=8
By Lagrange's Theorem, the possible orders are 1,2,4, and 8
The only subgroup of order 1is {1} and the only subgroup of order 8 is D4
.If D4 has an order 2 subgroup, it must be isomorphic to Z2 (this is the only group of order 2 up to isomorphism). Such a group is cyclic, it is generated by an element of order 2.
If D4 has an order 4 subgroup, it must be isomorphic to either Z4 or Z2×Z2 (these are the only groups of order 4 up to isomorphism). In the former case, the group is cyclic, it is generated by an element of order 4.
In summary, first find all the elements of order 2
and all the elements of order 4; each of them
Proposition: Every subgroup of Dn is cyclic or dihedral. A complete listing of the subgroups (including 1 and Dn) is as follows:
(1) r^d for all divisors dn.
(2) r^d,r^i/s, where dn and 0id1.generates a cyclic subgroup. Then consider pairs of elements of order 2 to find subgroups isomorphic to Z2×Z2.
There are a list of subgroups of D4, three of which have order 3. These subgroups are
all the centralizers of the di erent elements of the group. The question is, are there
any others? The answer to that is no. The hard part is to see why are there no others.
Note that there are two elements of order 4, namely R90 and R270. They each
generate the same subgroup of order 4, which is on the list. All other elements of D4
have order 2. Also notice that all three subgroups of order 4 on the list contain R180,
which commutes with all elements of the group. Now let H =fe; a; b; cg be a subgroup
of order 4 not on the list. Thus the elements a; b; c all have order 2 (for if H contained
an element of order 4, it would equal a subgroup on the list). Furthermore, these
elements must all commute with each other,otherwise we would get a fth element.
To see why, suppose that ab = c, while ba 6= c, Then ba is a new element not already in
H. Hence all pairs of elements in H commute. Now one of a, b or c is not a rotation,
otherwise H =fR90;R180;R270;R0g which is on the list. Say a is not a rotation. Then
H = C(a), which again means H is on the list, and we are done.
(2) (3.22)
Recall that
D4 ={ e,R90, R180, R270, F, FR90, FR180, FR270.
There are three subgroups of order 4, one cyclic and two not:
<R90>=< R270>={ e,R90, R180, R270}
{e, F,R180, FR180}
{e,R180, FR90, FR270}.
(
(b)
Let G be a group of order n. Then every element of G has order dividing n.
Let g be a non-identity element with order dividing n, say m
If mis prime, then we are done.
If m is composite, write m=pq, where p is a prime.
Then g^q is an element of order p in G
For your proof, you need to relate finiteness of G
to the order of g. Also, you need to show |g^n/p|=p not just because (g^n/p)p=e.
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