Extensions and the irreducibilities of the induced characters of cyclic p-groups

Extensions and the irreducibilities of the induced characters of cyclic p-groups

Katsusuke Sekiguchi

(Received July 21, 2001) (Revised January 4, 2002)

Abstract. Letfbe a faithful irreducible character of the cyclic groupCnof order pⁿ, where pis an odd prime. We study the p-groupGcontainingCnsuch that the induced character f^G is also irreducible. The purpose of this paper is to determine the subgroupsN_GðNGðCnÞÞandN_GðNGðNGðCnÞÞÞofGin the case when½NGðCnÞ;C_n ¼p.

1. Introduction

Let G be a finite group. We denote by IrrðGÞ the set of complex irre- ducible characters of G and by FIrrðGÞ ðHIrrðGÞÞ the set of faithful irreduc- ible characters of G.

Let p be a prime. For a non-negative integer n, we denote by Cn the cyclic group of order pⁿ. A finite group G is called an M-group, if every fAIrrðGÞ is induced from a linear character of a subgroup of G.

It is well-known that every nilpotent group is an M-group. Hence, when Gis a p-group, for anywAIrrðGÞ, there exists a subgroupH ofG and a linear character f of H such that f^G¼w. If we set N¼Kerf, then NkH and f is a faithful irreducible character of H=NGC_n, for some non-negative integer n. In this paper, we will consider the case when N¼1, that is, f is a faithful linear character of HGC_n.

We consider the following:

Problem 1. Let p be an odd prime, and f be a faithful irreducible character of Cn. Determine the p-group G such that CnHG and the induced character f^G is also irreducible.

Since all the faithful irreducible characters of C_n are algebraically con- jugate to each other, the irreducibility of f^G ðfAFIrrðCnÞÞ is independent of the choice of f, and depends only on n.

2000 Mathematics Subject Classification. 20C15.

Key words and phrases. p-group, extension, irreducible induced character, faithful irreducible character.

This problem has been solved in each of the following cases:

(1) C_nkG ([2]),

(2) Ghas a subgroup H containingCn such thatCnkH and½G:H ¼p ([6]).

On the other hand, when p¼2, Yamada and Iida [4] proved the following interesting result:

Let Q denote the rational field. Let G be a 2-group and w a complex irreducible character of G. Then there exist subgroups HjN in G and a complex irreducible character f of H such that w¼f^G, QðwÞ ¼QðfÞ, N¼ Kerf and

H=NGQn ðnb2Þ; or Dn ðnb2Þ; or SDn ðnb3Þ; or Cn ðnb0Þ:

Here, Qn, Dn and SDn denote the generalized quaternion group, the dihedral group of order 2^nþ1 ðnb2Þand the semidihedral group of order 2^nþ1 ðnb3Þ, respectively, and QðwÞ ¼QðwðgÞ;gAGÞ.

They considered the following:

Problem 2. Let f be a faithful irreducible character of H, where H¼Qn

or D_n or SD_n. Determine the 2-group G such that HHG and the induced character f^G is also irreducible.

Yamada and Iida [3] solved this problem in the case when ½G;H ¼2 or 4 and we have solved it when ½G;H ¼8 ([5]) for all H¼Q_n or D_n or SD_n.

Moreover, we have recently solved Problem 2 completely ([7]). In [7], we showed that

G¼NGðHÞ or NGðNGðHÞÞ;

for all H¼Q_n orD_n orSD_n, if Gsatisfies the conditions of Problem 2. Here, as usual, N_GðHÞ and N_GðNGðHÞÞ are the normalizers of H and N_GðHÞ in G, respectively. This means that, if we define subgroups of G by

M1¼NGðHÞ; and Miþ1¼NGðMiÞ; for ib1;

then

HHM₁HM₂¼M₃ ¼M₄¼ ¼G;

for all H ¼Qn or Dn or SDn.

In this paper, we consider Problem 1. We also define subgroups ofG by N₁¼N_GðCnÞ; and Niþ1¼N_GðNiÞ; for ib1:

The purpose of this paper is to determine the groups N₂¼N_GðNGðCnÞÞ and N3¼NGðNGðNGðCnÞÞÞ in the case where ½NGðCnÞ;Cn ¼p and ½G;Cnbp³. As a consequence of the results, we will see that

CnYN1YN2YN3; in this case.

Throughout this paper, ZandNdenote the set of rational integers and the natural numbers, respectively. We will frequently use the word ‘‘respectively’’

so it is abbreviated to ‘‘resp.’’.

2. Statements of the results

For the rest of this paper, we assume that p is an odd prime. First, we introduce the following groups:

( i ) Gðn;mÞ ¼ha;b_mi with a^pn ¼b_m^pm ¼1, b_mab¹_m ¼a^1þpnm, ðman1Þ.

( ii ) Gðn;m;1Þ ¼ha;bm;vi ðjGðn;mÞ ¼ha;bmiÞwitha^pn¼b_m^pm¼1,bmab¹_m ¼ a^1þpnm, vav¹¼a^1þpnm1b_m^pm1, v^p¼b_m, vb_mv¹¼b_m ð2man1Þ.

(iii) Gðn;1;1;1Þ ¼ha;b₁;v;xi ðjGðn;1;1Þ ¼ha;b₁;viÞ with a^pn¼b₁^p¼1, b1ab¹₁ ¼a^1þpn1,vav¹¼a^1þpn2b1,v^p¼b1,vb1v¹¼b1,xax¹¼a^1þpn3v, x^p¼v, xvx¹¼v, xb₁x¹¼b₁ ð7anÞ.

We can see that Gðn;m;1Þ (resp. Gðn;1;1;1Þ) is an extension group of Gðn;mÞ (resp. Gðn;1;1Þ) by using Proposition 1 below:

Proposition1. Let N be a finite group such that GjN and G=N¼huNiis a cyclic group of order m. Then u^m¼cAN. If we put sðxÞ ¼uxu¹, xAN, then sAAutðNÞ and (i) s^mðxÞ ¼cxc¹; ðxANÞ (ii) sðcÞ ¼c.

Conversely, if sAAutðNÞand cAN satisfy (i)and (ii), then there exists one and only one extension group G of N such that G=N¼huNiis a cyclic group of order m and sðxÞ ¼vxv¹ ðxANÞ and v^m¼c.

Proof. For instance, see [8, III, § 7].

Theorem0.1 (Iida [2]). Let G be a p-group which contains C_n as a normal subgroup of index p^m. Let fAFIrrðCnÞ. Suppose that f^GAIrrðGÞ. Then man1, and GGGðn;mÞ.

In particular, when C_nHG and ½G:C_n ¼p, C_n is always a normal subgroup of G. Hence we have:

Corollary 0.1. Let fAFIrrðCnÞ. Suppose that a group G containing C_n satisfies ½G:Cn ¼p and f^GAIrrðGÞ. Then GGGðn;1Þ.

Theorem 0.2 ([6]). Let G be a p-group which contains Cn, and let fA FIrrðCnÞ. Suppose that ½G:C_n ¼p^mþ1, f^GAIrrðGÞ, and n3b2m. Fur- ther, suppose that there exists a subgroup H of G such that HjCn and

½G:H ¼p. Then

(1) GGGðn;mþ1Þ if Cn is a normal subgroup of G.

(2) GGGðn;m;1Þ if Cn is not a normal subgroup of G.

Corollary 0.2. Let G be a p-group which contains Cn and let fA FIrrðCnÞ. Suppose that ½G:C_n ¼p², f^GAIrrðGÞ and nb5. Then

(1) GGGðn;2Þ if Cn is a normal subgroup of G.

(2) GGGðn;1;1Þ if C_n is not a normal subgroup of G.

Our main theorem is the following:

Theorem. Let p be an odd prime. Let G be a p-group which contains C_n¼hai. We assume that ½G:C_nbp³. Define the subgroups of G by

N1¼NGðCnÞ; and Niþ1¼NGðNiÞ; for i¼1;2:

Let fAFIrrðCnÞ and 7an. Suppose that f^GAIrrðGÞ, and ½N1:Cn ¼p.

Then

(1) N2=N1GC1 and N2GGðn;1;1Þ, (2) N3=N2GC1 and N3GGðn;1;1;1Þ.

Remark 1. Conversely, it is easy to see that the groups Gðn;1;1Þ and Gðn;1;1;1Þ satisfy the condition ðEX;CÞ, which is defined in section 3 of this paper. Hence these groups satisfy the conditions of Problem 1.

Remark 2. By results of Iida ([2], see Theorem 0.1. in this paper), we can see that N1GGðn;1Þ.

3. Some preleminary results

In this section, we state some results concerning the criterion of the irre- ducibilities of induced characters and others, which we need in section 4.

We denote by z¼z_pⁿ a primitive pⁿth root of unity. It is known that, for Cn¼hai, there are pⁿ irreducible characters f_n ð1anapⁿÞ of Cn:

f_nðaⁱÞ ¼zⁿⁱ; ð1aiapⁿÞ:

The irreducible character f_n is faithful if and only if ðn;pÞ ¼1. It is well- known that

AuthaiGðZ=pⁿZÞ GC Cn1

where ðZ=pⁿZÞ is the unit group of the factor ringZ=pⁿZandC is the cyclic group of order p1. Further, Cn1 is generated by the element 1þp in Z=pⁿZ.

First, we state the following result of Shoda (cf [1, p. 329]):

Proposition 2. Let G be a group and H be a subgroup of G. Let f be a linear character of H. Then the induced character f^G of G is irreducible if

and only if, for each xAGH ¼ fgAGjgBHg, there exists hAxHx¹VH such that fðhÞ0fðxhx¹Þ. (Note that, when f is faithful, the condition fðhÞ0 fðxhx¹Þ holds if and only if h0xhx¹).

Using this result, we have the following:

Proposition 3. Let hai¼CnHG, and f be a faithful irreducible char- acter of Cn. Then the following conditions are equivalent:

(1) f^G is irreducible,

(2) For each xAGCn, there exists yA hai Vxhaix¹ such that xyx¹0y.

Definition. When the condition (2) of Proposition 3 holds, we say thatG satisfies (EX,C).

Let H be a group. For a normal subgroup N of H, and any g;hAH, we write

g1h ðmodNÞ

when g¹hAN. For an element gAH, we denote by jgj the order of g.

4. Proof of Theorem

Let fAFIrrðCnÞ. Since f^G¼ ðf^N1Þ^GAIrrðGÞ, we must have f^N1 A IrrðN1Þ. Therefore, by Corollary 0.1, we can take an element b1AN1Cn¼ fgAN₁jgBC_ng such that

N1¼ha;b1ja^pn¼b₁^p¼1;b1ab¹₁ ¼a^1þpn1iGGðn;1Þ:

Proof of (1). Since Gis a p-group and ½G:N1bp², by our assumption, we have

N1YNGðN1Þ ¼N2:

Take an elementvAN2N1¼ fgAN2jgBN1g such thatv^pAN1. Denote by N₁⁰ the subgroup of G generated by v and the elements of N₁. Then

½N₁⁰:N1 ¼p and N₁⁰jN1: Since C_n is not a normal subgroup of N₁⁰, we have

N₁⁰¼ha;b1;viGGðn;1;1Þ

by Corollary 0.2. Hence we may assume that the elementsa,b1, and vsatisfy the following relations:

a^pn¼b₁^p¼1; b₁ab¹₁ ¼a^1þpn1; vav¹¼a^1þpn2b₁;

v^p¼b₁; vb₁v¹¼b₁: ðIÞ

Hereafter, we write b instead of b₁ for the sake of simplicity.

Remark 1. More precisely, in [6], we have shown that there exist an integer s1;ðs1;pÞ ¼1 and vAN₁⁰, such that the elements a1 ¼a^s1;b, and v satisfy the same relations as (I). Since hai¼ha^s1i, we can take a1 instead of a, and hence, we may assume that N₁⁰¼ha;b;vi and a;b and v satisfy the same relations as (I).

To prove the theorem, we need the following:

Lemma 1. For any integers i;j, the following equalities hold.

(i) ab1ba ðmodha^pn1iÞ.

(ii) ba^pb¹ ¼a^p. (iii) ðaⁱb^jÞ^p¼a^ip. (iv) va^p2v¹¼a^p2. (v) v^jav^j¼a^1þjpn2b^j.

(vi) ðaⁱv^jÞ^p1a^piv^pj ¼a^pib^j ðmodha^pn1iÞ.

(vii) ðaⁱv^jÞ^p2 ¼a^p2i.

Proof of Lemma 1. (i), (ii), (iii) and (iv) can be shown by direct cal- culations.

(v). Since nb7, by our assumption, we have va^pn2v¹¼a^pn2, by (iv).

Hence, v^jav^j¼a^1þjpn2b^j, for any jAZ.

(vi). By (i) and (v),

v^jaⁱv^j¼ ða^1þjpn2b^jÞⁱ1a^ið1þjpn2Þb^ij ðmodha^pn1iÞ: Using this relation repeatedly, we can get

ðaⁱv^jÞ^p1a^pia^ijpn2ð1þ2þþðp1ÞÞ

bijð1þ2þþðp1ÞÞ

v^pj ðmodha^pn1iÞ

¼a^pia^{ijpn2ðpðp1Þ=2Þ}bijðpðp1Þ=2Þ

v^pj 1a^piv^pj¼a^pib^j ðmodha^pn1iÞ;

since p is odd.

(vii) follows from (vi).

The assertion (1) follows from the following Claim I. N₁⁰¼N2.

Proof of Claim I. Suppose that N₁⁰YN₂. Take an element wA N2N₁⁰¼ fgAN2jgBN₁⁰g such that w^pAN₁⁰.

Write waw¹¼aⁱ⁰b^j0 for some i0;j0AZ, 0ai0apⁿ1, 0aj0ap1.

Then

wa^pw¹¼ ðaⁱ⁰b^j0Þ^p¼a^pi0 by Lemma 1 (iii). Therefore

w^p2a^pw^p2¼a^pip

2 0 :

Since w^p2AN1, we must have

i₀^p211 ðmod pⁿ¹Þ:

So,

i011 ðmod pⁿ³Þ:

Hence we can write as i0¼1þk0pⁿ³, and waw¹¼a^1þk0pn3b^j0

for some integer k₀. Since n3b4, by our assumption, we have va^pn3v¹¼a^pn3;

by Lemma 1 (iv). Hence

v^pj0waw¹v^pþj0¼v^pj0a^1þk0pn3b^j0v^pþj0¼a^{1þk0pn3þðpj0Þpn2};

by Lemma 1 (v). This means that v^pj0wAN1, which contradicts the hypoth- esis that wBN₁⁰. Hence the proof of Claim I is completed.

Proof of (2). Since G is a p-group and ½G:N₂bp, by our assumption, we have

N2YNGðN2Þ ¼N3:

Take an element yAN3N2 such that y^pAN2. Denote by N₂⁰ the subgroup of G generated by y and the elements of N₂. Then

½N₂⁰:N₂ ¼p and N₂⁰jN₂: First, we show the following

Claim II. We can write as

yay¹¼a^1þkpn3v^j; yby¹¼a^pn1db;

yvy¹¼a^pn2sb^dv;

for some k;j;d;sAZ, such that ðk;pÞ ¼ ðj;pÞ ¼1.

Proof of ClaimII. First, we consider the elements yby¹. By Lemma 1 (vi), jaⁱv^jjbp² when ðj;pÞ ¼1. So, we must have yby¹¼a^d0b^t0 for some d0;t0AZ. But

1¼yb^py¹¼ ða^d0b^t0Þ^p¼a^d0p; by Lemma 1 (iii). Therefore

d₀10 ðmod pⁿ¹Þ:

Hence, we may write d₀¼pⁿ¹d and

yby¹¼a^pn1db^t0; for some d AZ.

Next, consider the element yay¹. Since yBN2, we must have yay¹¼aⁱv^j;

for some jAZ, ðj;pÞ ¼1. By Lemma 1 (vi), ya^py¹¼ ðaⁱv^jÞ^p¼a^piþmpn1b^j for some mAZ. Since yay¹B hai, ya^py¹ B hai, and ya^p2y¹¼ ða^piþmpn1b^jÞ^p¼a^p2iA hai, we must have

a^p2i0a^p2; by the condition (EX,C). Therefore

iB h1þpⁿ²i;

where h1þpⁿ²i is the subgroup of ðZ=pⁿZÞ generated by 1þpⁿ². But y^pa^p2y^p¼a^p2ip and y^pAN₂¼Gðn;1;1Þ. Hence

i^pA h1þpⁿ²i: Thus we may write as i¼1þkpⁿ³ and

yay¹¼a^1þkpn3v^j; for some integers k;j such that ðk;pÞ ¼ ðj;pÞ ¼1.

Since n1b6, by our assumption, we have, ya^pn1y¹¼ ða^1þkpn3v^jÞ^pn1¼a^pn1:

by Lemma 1 (vii). Taking the conjugate of both sides of the equality,bab¹¼ a^1þpn1, b y y, we get

ða^pn1db^t0Þða^1þkpn3v^jÞða^pn1db^t0Þ¹¼ ða^1þkpn3v^jÞa^pn1: Hence,

a^{ð1þkpn3Þð1þt0pn1Þ}v^j¼a^1þkpn3þpn1v^j:

Therefore,

t011 ðmod pÞ;

and hence

yby¹¼a^pn1db:

Finally, we consider the element yvy¹. Write yvy¹¼a^s0v^h0. Then yby¹¼yv^py¹¼ ða^s0v^h0Þ^p¼a^ps0þepn1b^h0;

for some integer e, by Lemma 1 (vi). Therefore a^pn1db¼a^ps0þepn1b^h0; and, we have

h₀11 ðmod pÞ; and s₀10 ðmod pⁿ²Þ:

Write

h0¼1þpl and s0 ¼pⁿ²s;

for some l;sAZ. Then we have

yvy¹¼a^pn2sv^1þpl ¼a^pn2sb^lv:

Taking the conjugate of both sides of the equality, vav¹¼a^1þpn2b by y, we get

ða^pn2sb^lvÞða^1þkpn3v^jÞða^pn2sb^lvÞ¹¼ ða^1þkpn3v^jÞa^pn2ða^pn1dbÞ;

since ya^pn2y¹¼a^pn2. Hence, we have

a^{1þkpn3þpn2þpn1l}bv^j¼a^{1þkpn3þpn2þpn1d}bv^j: Therefore,

d1l ðmod pÞ; Thus the proof of Claim II is completed.

Now, we consider the element y^p ðAN2¼Gðn;1;1Þ ¼ha;b;viÞ. Write y^p¼ a^r0v^h. Then

a^r0v^h¼y^p¼yy^py¹¼yða^r0v^hÞy¹¼ ða^1þkpn3v^jÞ^r0ða^pn2sb^dvÞ^h: Therefore

v^h1v^jr0v^h ðmodN₁¼ha;biÞ:

Since ðj;pÞ ¼1, we have r010 ðmod pÞ. Hence we can write as r0¼pr, and

y^p¼a^prv^h; for some rAZ.

We show the following:

Claim III. There exists an integer e such that ða^eyÞ^p¼v^pt0þh, for some integer t0. Further, ðh;pÞ ¼1.

Proof of Claim III. It is easy to see that ha^pn3i is a normal subgroup of N₂⁰. By Claim II, the following equalities hold:

yay¹1av^j ðmodha^pn3iÞ;

yby¹1b ðmodha^pn3iÞ;

yvy¹1b^dv ðmodha^pn3iÞ:

Using these relations repeatedly, we can get

y^pay^p1abdjð1þ2þþðp1ÞÞv^pj¼ab^{djðpðp1Þ=2Þ}v^pj ¼av^pj¼ab^j ðmodha^pn3iÞ;

since p is odd. Hence we can write as

y^pay^p¼a^1þbpn3b^j;

for some bAZ. On the other hand, since y^p¼a^prv^h, we have y^pay^p¼ ða^prv^hÞaða^prv^hÞ¹¼a^1þhpn2b^h;

by Lemma 1 (v). Thus we have b^j ¼b^h, and, in particular, ðh;pÞ ¼1.

Define the subgroup H of N₂⁰ as

H ¼ha^pn3ihbi:

It is easy to see that N₂⁰jH, and the following equalities hold:

va1av ðmodHÞ; yay¹1av^j ðmodHÞ;

yvy¹1v ðmodHÞ:

Using these relations repeatedly, we can get

y^ca^my^c1a^mv^mcj ðmodHÞ; for any m;cAZ. Using this equality, we have

ða^myÞ^p1a^pmvjmð1þ2þþðp1ÞÞy^p¼a^pmv^{jmðpðp1Þ=2Þ}a^prv^h¼a^pðmþrÞv^h ðmodHÞ; for any mAZ. Therefore we may write as

ða^myÞ^p ¼a^{pðmþrÞþypn3}b^tv^h;

for some integers y and t. Note that y and t are not independent of the choice of m. If we set y₁¼a^ry, then

y₁^p¼a^y0pn3b^t0v^h¼a^y0pn3v^pt0þh;

for some integers y0 and t0. Further, set e¼ y0pⁿ⁴r, and y2¼a^ey¼a^y0pn4ry¼a^y0pn4y1:

Since n4b3, by our assumption, we have y1a^pn4y¹₁ ¼a^pn4: Hence,

y₂^p¼ ða^eyÞ^p¼ ða^y0pn4y1Þ^p¼a^y0pn3y₁^p¼v^pt0þh: This completes the proof of Claim III.

Since ðpt0þh;pÞ ¼1, there exists k⁰AZ, such that ðpt0þhÞk⁰11 ðmodp²Þ. Hence

y₂^k0p ¼v^ðpt0þhÞk0¼v:

Therefore

y2vy¹₂ ¼v; and y2by¹₂ ¼b:

Further, we have

y2ay¹₂ ¼a^y0pn4ryay¹a^y0pn4þr

¼a^rða^1þkpn3v^jÞa^r

¼a^ra^1þkpn3ða^1þjpn2b^jÞ^rv^j

1a^ra^1þkpn3a^rb^jrv^j ðmodha^pn2iÞ 1a^1þkpn3b^jrv^j ðmodha^pn2iÞ

¼a^1þkpn3v^pjrþj: Therefore

y₂ay¹₂ ¼a^{1þkpn3þdpn2}v^pjrþj;

for some integer d. If we set k₁ ¼kþdp, then ðk1;pÞ ¼1, and y2ay¹₂ ¼a^1þk1pn3v^pjrþj:

Since n3b4, by our assumption, we have y2a^pn3y¹₂ ¼a^pn3: Thus,

y₂^pay^p₂ ¼a^1þk1pn2v^pj¼a^1þk1pn2b^j: On the other hand, since y₂^p¼v^pt0þh, we have

y₂^pay^p₂ ¼ ðv^pt0þhÞaðv^pt0þhÞ¹¼a^{1þðpt0þhÞpn2}b^pt0þh¼a^{1þðpt0þhÞpn2}b^h; by Lemma 1 (v). Therefore we have

k11pt0þh ðmod p²Þ;

and

j1h ðmodpÞ:

Summarizing the results, we have

y₂ay¹₂ ¼a^1þk1pn3v^jprþj¼a^1þk1pn3b^hrv^j; y₂^p¼v^k1;

y2by¹₂ ¼b;

y2vy¹₂ ¼v:

There exists an integer l1, such that

l1k111 ðmodp³Þ:

Since

k11h1j ðmod pÞ;

we have

l₁k₁1l₁j11 ðmod pÞ:

Hence, we may write as

l₁j¼1þps₂; for some s₂AZ. Set y₃ ¼y₂^l1. Then

y₃ay¹₃ ¼y₂^l1ay^l₂ ¹ ¼a^1þpn3k1l1v^l1ðjprþjÞ

¼a^1þpn3v^l1jðprþ1Þ¼a^1þpn3v^{ðps2þ1Þðprþ1Þ}

¼a^1þpn3v^{pðrþs2Þþ1}¼a^1þpn3b^r1v;

where r1¼rþs2. Further

y₃^p¼y₂^pl1¼v^k1l1 ¼v; y3vy¹₃ ¼v; and y3by¹₃ ¼b:

If we set a0¼a^1pr1 ð¼a^1þðpnpÞr1Þ, then we have a₀^pn¼1 and ba0b¹¼a₀^1þpn1: Further, we have

va0v¹¼va^1pr1v¹¼vaa^pr1v¹¼ ða^1þpn2bÞða^1þpn2bÞ^pr1

¼ ða^1þpn2bÞa^{ð1þpn2Þðpr1Þ}¼a^{ð1þpn2Þð1pr1Þ}b¼a₀^1þpn2b;

by using Lemma 1 (iii). We also have y₃a₀y¹₃ ¼y₃aa^pr1y¹₃

¼ ða^1þpn3b^r1vÞða^1þpn3b^r1vÞ^pr1

1ða^1þpn3b^r1vÞða^pð1þpn3Þv^pÞ^r1 ðmodha^pn1iÞ

¼ ða^1þpn3b^r1vÞa^{pr1ð1þpn3Þ}b^r1;

1a^1þpn3a^{pr1ð1þpn3Þ}v ðmodha^pn1iÞ

¼a^{ð1þpn3Þð1pr1Þ}v¼a₀^1þpn3v:

Hence we may write as y₃a₀y¹₃ ¼a₀^1þpn3a^gpn1v, for some integer g. But a₀^gpn1 ¼a^gpn1, so

y₃a₀y¹₃ ¼a₀^1þpn3þgpn1v:

Finally, we set y₄¼b^gy₃. Then we have

y₄^p¼y₃^p¼v; y₄vy¹₄ ¼v; and y₄by¹₄ ¼b:

Further

y4a0y¹₄ ¼b^gy3a0y¹₃ b^g¼b^gða₀^1þpn3þgpn1vÞb^g

¼a^ð1þp₀ ^{n3þgpn1Þð1þpn1Þg}v¼a₀^1þpn3v:

Therefore, the relations of the elements a0;b;v, and y4 are the same as that of Gðn;1;1;1Þ. So, the group N₂⁰¼ha₀;b;v;y₄i is clearly isomorphic to Gðn;1;1;1Þ.

We will complete the proof of (2), by showing the following:

Claim IV. N₂⁰¼N3.

Proof of Claim IV. Since the proof of this claim is similar to that of Claim I, we state only an outline of the proof. Suppose that N₂⁰YN₃. Take an element gAN3N₂⁰ such that g^pAN₂⁰. Then we can write as

ga0g¹¼a₀^1þk2pn4v^j1

for some integers k2 and j1, by the same way as in the proof of Claim I.

Since n4b3, by our assumption, we have y4a₀^pn4y¹₄ ¼a₀^pn4: Thus

y₄^p2j1ga0g¹y^p₄ ^2þj1 ¼a₀^{1þk2pn4þðp2j1Þpn3}:

This means that y₄^p2j1gAN1, which contradicts the hypothesis that gA N3N₂⁰. Hence the proof of Claim IV is completed.

References

[ 1 ] C. Curtis and I. Reiner: ‘‘Representation theory of finite groups and associative algebras’’, Interscience, New York, 1962.

[ 2 ] Y. Iida: Thep-groups with an irreducible character induced from a faithful linear character, Preprint.

[ 3 ] Y. Iida and T. Yamada: Extensions and induced characters of quaternion, dihedral and semidihedral groups, SUT J. Math. 27 (1991), 237–262.

[ 4 ] Y. Iida and T. Yamada: Types of faithful metacyclic 2-groups, SUT J. Math.28(1992), 23–46.

[ 5 ] K. Sekiguchi: Extensions of some 2-groups which preserve the irreducibilities of induced characters, Osaka J. Math. 37(2000).

[ 6 ] K. Sekiguchi: Irreducibilities of the induced characters of cyclic p-groups, Math. J. of Okayama Univ. 41 (1999).

[ 7 ] K. Sekiguchi: Extensions and the irreducibilities of induced characters of some 2-groups, Hokkaido Math. J. 31 (2002).

[ 8 ] H. Zassenhaus: ‘‘The theory of groups’’, Chelsea, New York, 1949.

Department of Civil Engineering Faculty of Engineering

Kokushikan University 4-28-1 Setagaya Setagaya-Ku

Tokyo 154-8515, Japan